case study chapter 9 class 10 maths

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

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Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
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Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Chapter 9 Class 10 Some Applications of Trigonometry

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Class 10 Maths Case Study Questions PDF Download

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Are you looking for a reliable source to download Class 10 Maths case study questions in PDF format? Look no further! In this article, we will provide you with a comprehensive collection of case study questions specifically designed for Class 10 Maths Case Study Questions . Whether you are a student or a teacher, these case study questions will prove to be a valuable resource in your preparation or teaching process.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 10  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

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CBSE 10th Maths: Case Study Questions With Answers

Students taking the 10th board examinations will see new kinds of case study questions in class. The board initially incorporated case study questions into the board exam. The chapter-by-chapter case study question and answers are available here.

Chapterwise Case Study Questions for Class 10 Mathematics

Case study questions are an essential component of the Class 10 Mathematics curriculum. They provide students with real-world scenarios where they can apply mathematical concepts and problem-solving skills. By analyzing and solving these case study questions, students develop a deeper understanding of the subject and improve their critical thinking abilities.

The above  Case studies for Class 10 Maths  will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of schools.studyrate.in for the benefit of Class 10 students.

• Class 10th Science Case Study Questions

Benefits of Case Study Questions for Class 10 Mathematics

Case study questions offer several benefits to both students and teachers. Here are some key advantages:

• Practical Application : Case study questions bridge the gap between theory and real-life situations, allowing students to apply mathematical concepts in practical scenarios.
• Analytical Thinking : By solving case study questions, students enhance their analytical thinking and problem-solving skills.
• Conceptual Clarity : Case study questions help reinforce the fundamental concepts of mathematics, leading to improved conceptual clarity.
• Exam Preparation : Practicing case study questions prepares students for their Class 10 Mathematics exams, as they become familiar with the question formats and types.
• Comprehensive Assessment : Teachers can use case study questions to assess students’ understanding of various mathematical concepts in a comprehensive manner.

How to Use Case Study Questions Effectively

To make the most out of the case study questions, follow these effective strategies:

• Read the question carefully : Understand the given scenario and identify the mathematical concepts involved.
• Analyze the problem : Break down the problem into smaller parts and determine the approach to solve it.
• Apply relevant formulas and concepts : Utilize your knowledge of the subject to solve the case study question.
• Show your working : Clearly demonstrate the steps and calculations involved in reaching the solution.
• Check your answer : Always verify if your solution aligns with the given problem and recheck calculations for accuracy.

Tips for Solving Case Study Questions

Here are some useful tips to excel in solving case study questions:

• Practice regularly : Regular practice will enhance your problem-solving skills and familiarity with different question formats.
• Understand the concepts: Ensure you have a strong foundation in the underlying mathematical concepts related to each chapter.
• Work on time management : Practice solving case study questions within a stipulated time to improve your speed and efficiency during exams.
• Seek clarification : If you encounter any doubts or difficulties, don’t hesitate to seek guidance from your teacher or peers.

Case study questions are an invaluable resource for Class 10 Mathematics students. They provide practical application opportunities and strengthen conceptual understanding. By utilizing the chapter-wise case study questions provided in this article, students can enhance their problem-solving skills, prepare effectively for exams, and develop a deeper appreciation for the subject.

FAQs on Class 10 Maths Case Study Questions

Q1: can i download the class 10 maths case study questions in pdf format.

Yes, you can download the Class 10 Maths case study questions in PDF format from our site free of cost.

Q2: Are the case study questions aligned with the latest curriculum?

Yes, the case study questions presented in this article are designed to align with the latest Class 10 Mathematics curriculum.

Q3: How can case study questions improve my exam preparation?

Case study questions help you understand the practical application of mathematical concepts, enabling you to approach exam questions with greater confidence and clarity.

Q5: Where can I find more resources for Class 10 Mathematics preparation?

Download more resources of Class 10th Maths from schools.studyrate.in, we offer additional resources and practice materials for Class 10 Mathematics.

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Case Study Questions Class 10 Maths with Solutions PDF Download

 case study questions class 10 maths pdf, case study questions class 10 maths with solutions, case study questions class 10 maths cbse chapter wise pdf download, how to solve case-based question in maths.

• First of all, a student needs to read the complete passage thoroughly. Then start solving the question
• After reading the question try to understand from which topics the question is asked. and try to remember all the concepts of that topic.
• Sometimes the question is very tricky and you will find it very difficult to understand. In that case, Read the question and passage again and again.
• After solving the answer check your answer with the options given.
• Remember, write only answering your answer book

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Case Study Class 10 Maths Questions

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Case Study Class 10 Maths Questions and Answers (Download PDF)

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study. 

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them. 

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers. 

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

Features Of Class 10 Maths Case Study Questions And Answers Pdf

Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-

• Accurate answers of all the Case-based questions given in the PDF.
• Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
• Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.

Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers

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Important Case Study Questions CBSE Class 10 Maths

Cbse class 10 maths important case study questions: cbse class 10 maths exam 2024 is just around the corner. case study questions can be a hard nut to track if not prepared well. check here important case study questions from class 10th maths curriculum  for cbse class 10 maths board exam 2024..

CBSE Class 10 Maths Question Paper Structure

• Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 
• Section B : 5 Short Answer-I (SA-I) type questions carrying 2 marks each. 
• Section C : 6 Short Answer-II (SA-II) type questions carrying 3 marks each. 
• Section D : 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E : 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.

CBSE Class 10 Maths Important Case Study Questions

Related:  CBSE Class 10 Maths Important Formulas for Last Minute Revision for Board Exam 2024

1 Two Friends Geeta and Sita were playing near the river. So, they decide to play a game in which they have to throw the stone in the river, and whoever will throw the stone at maximum distance, win the game. Geeta Starts first and throws the stone in the river. During her throw, her hand was making an angle of 60° with the Horizontal plane. Sita throws at 45°.

• Straight Line
• Semi circle
• Bi-Quadratic
• Parabola Open Upward
• Parabola Open Downward
• Hyperbola Open Upward
• Hyperbola Open downward
• Two Real Points
• One Real Point
• Three Real Points
• Putting y=0 in given Polynomial
• Putting y=1 in the given Polynomial
• Putting x=0 in the given Polynomial.
• Putting x=1 in the given Polynomial.

2 The department of Computer Science and Technology is conducting an International Seminar. In the seminar, the number of participants in Mathematics, Science and Computer Science are 60, 84 and 108 respectively. The coordinator has made the arrangement such that in each room, the same number of participants are to be seated and all of them being in the same subject. Also, they allotted the separate room for all the official other than participants.

(i) Find the total number of participants.

(a) 60 

(b) 84 

(c) 108 

(d) none of these

(ii) Find the LCM of 60, 84 and 108.

(a) 12 

(b) 504 

(c) 544320 

(iii) Find the HCF of 60, 84 and 108.

(iv) Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

(b) 20 

(c) 21 

(v) Based on the above (iv) conditions, find the minimum number of rooms required for all the participants and officials.

• In the standard form of quadratic polynomial, ax2 + bx + c, what are a, b and c ?
• If the roots of the quadratic polynomial are equal, what is the discriminant D ?
• If α and 1/α are the zeroes of the quadratic polynomial 2x2 – x + 8k, then find the value of k ?
• Represent the first situation algebraically.

a) 12x+10y=11200

b) 10x+12y=11200

c) 12x-10y=11200

d) 10x-12y=1120

2 Represent the second situation algebraically

a) 46x+55y=51400

b) 55x+46y=51400

c) 55x-46y=51400

d) 46x-55y=51400

3 The system of linear equations representing both the situations will have.

a) Infinite number of solutions

b) Unique solution

c) No Solutions

d) Exactly two solutions

4 The graph of the system of linear equations representing both the situations will be

a) Parallel lines

b) Coincident lines

c) Intersecting lines

d) None of these

• Represent algebraically the situation in hall “Rose”.

a) 50x + y = 10000

b) 50x − y = 10000

c) x + 50y = 10000

d) x − 50y = 10000

2 Represent algebraically the situation in hall “Jasmine”

a) x + 25y = 7500

b) x − 25y = 7500

c) 25x + y = 7500

d) 25x − y = 7500

3 What is the fixed rent of the halls?

4 Find the amount the hotel charged per person.

6 Riya has a field with a flowerbed and grassland. The grassland is in the shape of a rectangle while the flowerbed is in the shape of a square. The length of the grassland is found to be 3 m more than twice the length of the flowerbed. Total area of the whole land is 1260m2

(a)If the length of the square is x m then find the total length of the field i

(b) What will be the perimeter of the whole figure in terms of x?

(c )Find the value of x if the area of total field is 1260 m2

(d) Find the area of grassland and the flowerbed separately.

7 Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do it in 31 seconds.

1 Write first four terms are in AP for the given situation.

2 What is the minimum number of days he needs to practice till his goal is achieved.

3 How many seconds it takes after the 5th day .

8 Helicopter Patrolling: A helicopter is hovering over a crowd of people watching a police standoff in a parking garage across the street. Stewart notices the shadow of the helicopter is lagging approximately 57 m behind a point directly below the helicopter. If he is 160 cm tall and casts a shadow of 38 cm at this time,

(i) what is the altitude of the helicopter?

(ii) What will be length of shadow of Stewart at 12:00 pm

(iii) Write the name of triangles formed for this situation.

9 Seema has a 10 m × 10 m kitchen garden attached to her kitchen. She divides it into a 10 ×10 grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sow a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kusum visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully and answer the following questions:

(i) Find the distance between A and B is

(ii) Find the mid- point of the distance AB

(iii) Find the distance between B and C

10 A heavy-duty ramp is used to winch heavy appliances from street level up to a warehouse loading dock. If the ramp is 2 meter high and the incline is 4 meter long.

(Use √3 = 1.73)

a What angle does the dock make with the street?

b How long is the base of the ramp? ( In round figure)

• If the length of the base is 12 cm and the height is 5 cm then the length of the hypotenuse of that sandwich is:
• If he increases the base length to 15 cm and the hypotenuse to 17 cm, then the height of the sandwich is :
• The value of tan 45° + cot 45°

(a) 1 (b) 2 (c) 3 (d) 4

12 A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°.

(i) the height of the pole

(ii) radius( height) of the dome

(iii) Is it possible to see the pole at the angle of 60 0

(iv) If the height of pole is increased, the angle elevation will .....

14 John had a farm with many animals like cows, dogs, horses etc. He had sufficient grass land for the cows and horses to graze, One day Three of his horses were tied with 7 metre long ropes at the three corners of a triangular lawn having sides 20m, 34m and 42m.

(a) Find the area of the triangular lawn .

(b) Find the area of the field that can be grazed by the horses.

(c) The area that cannot be grazed by the horses.

15 Arun, a 10th standard student, makes a project on coronavirus in science for an exhibition in his school. In this project, he picks a sphere which has volume 38808 cm3 and 11 cylindrical shapes, each of volume 1540 cm3 with length 10 cm.

Based on the above information, answer the following questions.

(i) Diameter of the base of the cylinder is

(ii) Diameter of the sphere is

(iii) Total volume of the shape formed is

(iv) Curved surface area of the one cylindrical shape is

(v) Total area covered by cylindrical shapes on the surface of sphere is

• In which age group, will the maximum number of children belong?
• Find the mode of the ages of children playing in the park?

17 Piggy bank or Money box( a coin container) is normally used by children. Piggy bank serves as a pedagogical device to teach about saving money to children. Generally, piggy banks have openings besides the slot for inserting coins but some do not have openings. We have to smash the piggy bank with a hammer or by other means, to get the money inside it. A child Shreya has a Piggybank. She saves her money in her Piggybank. One day she found that her Piggybank contains hundred 50 paisa coins, fifty 1 rupees coin, twenty 2 rupees coin, and ten 5 rupees coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down.

(a) The probability that the fallen coin will be 50 paisa coin, is -------

(b) The probability that the fallen coin will be 5 rupees coin, is---------

(c) The probability that the fallen coin will be 2 rupees coin, is---------

(d) The probability that the fallen coin will be 2 rupees coin or 5 rupees coin, is--------

• Find the probability of getting no heads
• Find the probability of getting one tail

Download answers to CBSE Class 10 Maths Important Case Study Questions

Chapter-wise important case study questions cbse class 10 maths.

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NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

Get Free NCERT Solutions for Class 10 Maths Chapter 9 Ex 9.1 PDF. Some Applications of Trigonometry Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 9.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 9 Maths Class 10 Some Applications of Trigonometry Exercise 9.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 9 Some Applications of Trigonometry:

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex Ex 9.1 are part of NCERT Solutions for Class 10 Maths . Here we have given NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1.

• Some Applications of Trigonometry Class 10 Ex 9.1
• प्रश्नावली 9.1 का हल हिंदी में
• Applications of Trigonometry Class 10 Extra Questions

Hi all, We have also solved 68 questions of Chapter 12 – Some Applications of Trigonometry from RD Sharma Class 10 Maths textbook. You can download these solutions in PDF from the above link.

You can also download the free PDF of  Ex 9.1 Class 10 Some Applications of Trigonometry NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 9 Some Applications of Trigonometry PDF

Class 10 Maths Some Application Of Trigonometry Mind Maps

Some application of trigonometry, introduction.

The height or length of an object or the distance between two distinct objects can be determined with the help of trigonometric ratios.

Line of Sight and Angle of Elevation

Angle of Depression

To find BC, we will use trigonometric ratios of ∠BAC or ∠A. In ∆ABC, the side BC is the opposite side to the known ∠A. Now we use either tan A or cot A, as these trigonometric ratios involve AB and BC to find BC. Therefore, tan A = \(\frac{B C}{A B}\) or cot A = \(\frac{A B}{B C}\), which on solving would give us BC. By adding AE to BC, you will get the height of the pole.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry (Hindi Medium) Ex 9.1

NCERT Solutions for Class 10 Maths

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

We hope the NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1, drop a comment below and we will get back to you at the earliest.

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CBSE 10th Standard Maths Subject Case Study Questions With Solution 2021 Part - II

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) Proportional expense for each person is

(iii) The fixed (or constant) expense for the party is

(iv) If there would be 15 guests at the lunch party, then what amount Mr Jindal has to pay?

(v) The system of linear equations representing both the situations will have

(ii) Represent the situation faced by Suman, algebraically

(iii) The price of one Physics book is

(iv) The price of one Mathematics book is

(v) The system of linear equations represented by above situation, has

(ii) Represent algebraically the situation of day- II.

(iii) The linear equation represented by day-I, intersect the x axis at

(iv) The linear equation represented by day-II, intersect the y-axis at

(v) Linear equations represented by day-I and day -II situations, are

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

(ii) Difference of pairs of shoes in 17 th  row and 10 th row is

(iii) On next day, she arranges x pairs of shoes in 15 rows, then x =

(iv) Find the pairs of shoes in 30 th row.

(v) The total number of pairs of shoes in 5 th and 8 th row is

(ii) The number on first card is

(iii) What is the number on the 19 th card?

(iv) What is the number on the 23 rd card?

(v) The sum of numbers on the first 15 cards is 

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information, answer the following questions. (i) Which of the following sequence is an A.P.?

(ii) If x, y and z are in A.P., then

(iii) If a 1  a 2 , a 3  ..... , a n are in A.P., then which of the following is true?

(iv) If the n th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(v) Which of the following is incorrect about A.P.?

(ii) Find the radius of the core.

(iii) S 2 =

(iv) What is the diameter of roll when one tissue sheet is rolled over it?

(v) Find the thickness of each tissue sheet

(ii) Distance travelled by aeroplane towards west after   \(1 \frac{1}{2}\)   hr is

(iii) In the given figure, \(\angle\) POQ is 

(iv) Distance between aeroplanes after  \(1 \frac{1}{2}\)   hr is

(v) Area of \(\Delta\) POQ is

(ii) The value of x is

(iii) The value of PR is 

(iv) The value of RQ is 

(v) How much distance will be saved in reaching city Q after the construction of highway? 

(ii) Length of BC =

(iii) Length of AD =

(iv) Length of ED = 

(v) Length of AE = 

(ii) The value of x + y is 

(iii) Which of the following is true?

(iv) The ratio in which B divides AC is

(v) Which of the following equations is satisfied by the given points?

(ii) The value of x is equal to

(iii) If M is any point exactly in between city A and city B, then coordinates of M are

(iv) The ratio in which A divides the line segment joining the points O and M is

(v) If the person analyse the petrol at the point M(the mid point of AB), then what should be his decision?

(ii) The centre of circle is the

(iii) The radius of the circle is

(iv) The area of the circle is

(v) If  \(\left(1, \frac{\sqrt{7}}{3}\right)\)   is one of the ends of a diameter, then its other end is

(ii) The distance between A and Cis

(iii) If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B?

(iv) If the fare for first bus is Rs10/km, then what will be the fare for total journey by that bus?

(v) If the fare for second bus is Rs 15/km, then what will be the fare to reach to the destination by this bus?

*****************************************

Cbse 10th standard maths subject case study questions with solution 2021 part - ii answer keys.

(i) (a): 1 st situation can be represented as x + 7y = 650 ...(i) and 2 nd situation can be represented as x + 11y = 970 ...(ii) (ii) (b): Subtracting equations (i) from (ii), we get  \(4 y=320 \Rightarrow y=80\) \(\therefore\)  Proportional expense for each person is Rs 80. (iii) (c): Puttingy = 80 in equation (i), we get x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90 \(\therefore\)  Fixed expense for the party is Rs 90 (iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290 (v) (a): We have a 1  = 1, b 1  = 7, c 1  = -650 and  \(a_{2}=1, b_{2}=11, c_{2}=-970 \) \(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\) \(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus, system of linear equations has unique solution.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850 (ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900 (iii) (c) : We have 2x + 3y = 850 .........(i) and 3x + 2y = 900 .........(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get 5y = 750 \(\Rightarrow\)   Y = 150 Thus, price of one Physics book is Rs 150. (iv) (d): From equation (i) we have, 2x + 3 x 150 = 850 \(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200 Hence, cost of one Mathematics book = Rs 200 (v) (a): From above, we have \(a_{1} =2, b_{1}=3, c_{1}=-850 \) \(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\) \(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus system of linear equations has unique solution.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600. (ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500. (iii) (c) : At x-axis, y = 0 \(\therefore\)   At y = 0, 2x + y = 1600 becomes 2x = 1600 \(\Rightarrow\) x = 800 \(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0). (iv) (d) : At y-axis, x = 0 \(\therefore\) 2x + Y = 1500 \(\Rightarrow\)  y = 1500 \(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500). (v) (b): We have, 2x + y = 1600 and 2x + y = 1500 Since  \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\) \(\therefore\) System of equations have no solution. \(\therefore\) Lines are parallel.

(i) (b): We have  \(6 x^{2}+x-2=0\) \(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \) \(\Rightarrow \quad(3 x+2)(2 x-1)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\) (ii) (c):  \(2 x^{2}+x-300=0\) \(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \) \(\Rightarrow \quad(x-12)(2 x+25)=0 \) \(\Rightarrow \quad x=12, \frac{-25}{2}\) (iii) (d):   \(x^{2}-8 x+16=0\) \(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\) (iv) (d):   \(6 x^{2}-13 x+5=0\) \(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \) \(\Rightarrow \quad(2 x-1)(3 x-5)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\) (v) (a):  \(100 x^{2}-20 x+1=0\) \(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)  

Number of pairs of shoes in 1 st , 2 nd , 3 rd row, ... are 3,5,7, ... So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2 (i) (d): Let n be the number of rows required. \(\therefore S_{n}=120 \) \(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \) \(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\) \(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\) So, 10 rows required to put 120 pairs. (ii) (b): No. of pairs in 1ih row = t 17 = 3 + 16(2) = 35 No. of pairs in 10th row = t 10  = 3 + 9(2) = 21 \(\therefore\) Required difference = 35 - 21 = 14 (iii) (c) : Here n = 15 \(\therefore\) t 15  = 3 + 14(2) = 3 + 28 = 31 (iv) (a): No. of pairs in 30 th row = t 30 = 3 +29(2) = 61 (v) (c): No. of pairs in 5 th row = t 5  = 3 + 4(2) = 11 No. of pairs in 8 th row = t 8  = 3 + 7(2) = 17 \(\therefore\) Required sum = 11 + 17 = 28

Let the numbers on the cards be a, a + d, a + Zd, ... According to question, We have (a + 5d) + (a + 13d) = -76 \(\Rightarrow\) 2a+18d = -76 \(\Rightarrow\) a + 9d= -38 ... (1) And (a + 7d) + (a + 15d) = -96 \(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2) From (1) and (2), we get 2d= -10 \(\Rightarrow\) d= -5 From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7 (i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5 (ii) (d): Number on first card = a = 7 (iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83 (iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103 (v) (d):  \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c) (ii) (c) (iii) (d) (iv) (b) (v) (c)

Here S n = 0.1n 2 + 7.9n (i) (c): S n -1 = 0.1(n - 1) 2 + 7.9(n - 1) = 0.1n 2 + 7.7n - 7.8 (ii) (b): S 1 = t 1  = a = 0.1(1) 2 + 7.9(1) = 8 cm = Diameter of core So, radius of the core = 4 cm (iii) (a): S 2 = 0.1(2) 2 + 7.9(2) = 16.2 (iv) (d): Required diameter = t 2 = S 2 - S 1 = 16.2 - 8 = 8.2 cm (v) (c): As d = t 2 - t 1  = 8.2 - 8 = 0.2 cm So, thickness of tissue = 0.2 \(\div\)   2 = 0.1 cm = 1 mm

(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\)  Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time =  \(\frac{3}{2}\)  hr. \(\therefore\)  Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after  \(1\frac{1}{2}\)   hour  \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of  \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\)   [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (b): If \(\Delta\) AED and \(\Delta\) BEC, are similar by SAS similarity rule, then their corresponding proportional sides are  \(\frac{B E}{A E}=\frac{C E}{D E}\) (ii) (c): By Pythagoras theorem, we have \(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\) (iii) (a): Since \(\Delta\) ADE and \(\Delta\) BCE are similar. \(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \) \(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\) (iv) (b): \(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \) \(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\) (v) (d) :   \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\) \(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \) \(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (a): We have, OA = 2 \(\sqrt{2}\) km \(\Rightarrow \sqrt{2^{2}+y^{2}}=2 \sqrt{2} \) \(\Rightarrow 4+y^{2}=8 \Rightarrow y^{2}=4 \) \(\Rightarrow y=2 \quad(\because y=-2 \text { is not possible })\) (ii) (c): We have OB = 8 \(\sqrt{2}\) \(\Rightarrow \sqrt{x^{2}+8^{2}}=8 \sqrt{2} \) \(\Rightarrow x^{2}+64=128 \Rightarrow x^{2}=64 \) \(\Rightarrow x=8 \quad(\because x=-8 \text { is not possible })\) (iii) (c) : Coordinates of A and Bare (2, 2) and (8, 8) respectively, therefore coordinates of point M are \(\left(\frac{2+8}{2}, \frac{2+8}{2}\right)\) i.e .,(5.5) (iv) (d): Let A divides OM in the ratio k: 1.Then \(2=\frac{5 k+0}{k+1} \Rightarrow 2 \mathrm{k}+2=5 k \Rightarrow 3 k=2 \Rightarrow k=\frac{2}{3}\) \(\therefore\) Required ratio = 2 : 3 (v) (b): Since M is the mid-point of A and B therefore AM = MB. Hence, he should try his luck moving towards B.

(i) (c): Required coordinates are  \(\left(0, \frac{4}{3}\right)\) (ii) (c) (iii) (a): Radius = Distance between (0,0) and  \(\left(\frac{4}{3}, 0\right)\) \(=\sqrt{\left(\frac{4}{3}\right)^{2}+0^{2}}=\frac{4}{3} \text { units }\) (iv) (b): Area of circle = \(\pi\) (radius) 2 \(=\pi\left(\frac{4}{3}\right)^{2}=\frac{16}{9} \pi \text { sq. units }\) (v) (d): Let the coordinates of the other end be (x,y). Then (0,0) will bethe mid-point of  \(\left(1, \frac{\sqrt{7}}{3}\right)\)  and (x, y). \(\therefore\left(\frac{1+x}{2}, \frac{\frac{\sqrt{7}}{3}+y}{2}\right)=(0,0) \) \(\Rightarrow \frac{1+x}{2}=0 \text { and } \frac{\frac{\sqrt{7}}{3}+y}{2}=0 \) \(\Rightarrow x=-1 \text { and } y=-\frac{\sqrt{7}}{3}\) Thus, the coordinates of other end be  \(\left(-1, \frac{-\sqrt{7}}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2). (i) (d): Required distance  \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\) \(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) (ii) (d): Required distance  \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\) \(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\) (iii) (b): Distance between Band C \(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\) Thus, distance travelled by first bus to reach to B \(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\) and distance travelled by second bus to reach to B \(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) \(\therefore\)  Distance of first bus is greater than distance of the second bus, therefore second bus should be chosen. (iv) (d): Distance travelled by first bus = 8.48 km \(\therefore\) Total fare = 8.48 x 10 = Rs 84.80 (v) (b): Distance travelled by second bus = 7. 2 km \(\therefore\) Total fare = 7.2 x 15 = Rs 108  

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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

NCERT Solutions Class 10 Maths Chapter 9 Some Applications of Trigonometry are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios. And you also learn what is the line of sight, angle of elevation, and angle of depression etc.  CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 9 Some Applications of Trigonometry Solutions

Below we have given the answers to all the questions present in Some Applications of Trigonometry in our NCERT Solutions for Class 10 Maths chapter 9. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

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Case Study Questions for Class 10 Maths

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This article covers case study questions for Class 10 Maths. Students are suggested to go through all the questions to score better in the exams.

Why Students Fear Case Study Questions for Class 10 Maths?

Students may fear Case Study questions for Class 10 Maths for the following reasons:

• Application of Concepts: Case Study questions require the application of mathematical concepts to real-world situations. Students may find it difficult to apply the concepts they have learned to solve the given scenario.
• Limited Practice: Unlike other types of questions, students may have limited practice with Case Study questions. This lack of familiarity may cause anxiety and fear among students.
• Lengthy and Complex Scenarios: Case Study questions often present lengthy and complex scenarios, which can be overwhelming for students. They may struggle to identify the relevant information and apply the appropriate mathematical concepts.
• Time Pressure: Case Study questions may require more time to solve compared to other types of questions. This can add to the stress and anxiety of students, especially during an exam.
• Fear of Making Mistakes: Since Case Study questions involve applying concepts to real-world situations, students may fear making mistakes or getting the wrong answer. This fear may cause them to avoid attempting the question altogether.

In conclusion, students may fear Case Study questions for Class 10 Maths due to the application of concepts, limited practice, lengthy and complex scenarios, time pressure, and fear of making mistakes. Teachers can help alleviate these fears by providing ample practice opportunities, breaking down the scenarios into smaller parts, and encouraging students to attempt the questions.

Best Way to Approach Case Study Questions for Class 10 Maths

Here are some tips on how to approach Case Study questions for Class 10 Maths:

• Read the scenario carefully: The first step is to read the scenario carefully and identify the key information. Pay attention to the given values, units, and any other important details.
• Identify the mathematical concepts involved: Once you have read the scenario, identify the mathematical concepts that are involved. This will help you determine which formulas or equations to apply.
• Break down the scenario into smaller parts: Some Case Study questions may have lengthy and complex scenarios. To make it easier, try to break down the scenario into smaller parts and identify the specific information that is needed to solve each part.
• Solve the problem step by step: Once you have identified the key information and the mathematical concepts involved, start solving the problem step by step. Show all the calculations and equations used.
• Check your answers: After you have solved the problem, check your answers to ensure that they are accurate and relevant to the scenario given. If possible, try to cross-check your answer using a different approach or formula.
• Practice, Practice, Practice: The more you practice Case Study questions, the more familiar you will become with the format and the types of scenarios presented. This will help you develop confidence and improve your performance.

In conclusion, approaching Case Study questions for Class 10 Maths involves careful reading of the scenario, identifying the mathematical concepts involved, breaking down the problem into smaller parts, solving the problem step by step, checking the answers, and practicing regularly.

Topics Covered in CBSE Class 10 Maths

Here are the topics covered in CBSE Class 10 Maths:

• Real Numbers: Euclid’s division lemma, HCF and LCM, irrational numbers, decimal representation of rational numbers, and the relationship between roots and coefficients of a quadratic equation.
• Polynomials: Zeros of a polynomial, relationship between zeros and coefficients of a polynomial, division algorithm for polynomials, and factorization of polynomials.
• Pair of Linear Equations in Two Variables: Graphical method of solution, algebraic methods of solution, and word problems based on linear equations.
• Quadratic Equations: Standard form of a quadratic equation, solutions of a quadratic equation by factorization and by using the quadratic formula, relationship between roots and coefficients, and nature of roots.
• Arithmetic Progressions: nth term of an AP, sum of first n terms of an AP, and word problems based on arithmetic progressions.
• Triangles: Properties of triangles, congruence of triangles, criteria for similarity of triangles, and Pythagoras theorem.
• Coordinate Geometry: Distance formula, section formula, area of a triangle, and equation of a line in different forms.
• Introduction to Trigonometry: Trigonometric ratios, trigonometric ratios of complementary angles, and word problems based on trigonometry.
• Some Applications of Trigonometry: Heights and distances.
• Circles: Tangent to a circle, number of tangents from a point on a circle, and chord properties.
• Constructions: Construction of bisectors of line segments and angles, construction of a triangle similar to a given triangle, and construction of a triangle of given perimeter and base angles.
• Areas Related to Circles: Areas of sectors and segments of a circle.
• Surface Areas and Volumes: Surface areas and volumes of spheres, cones, cylinders, and cuboids.

In conclusion, CBSE Class 10 Maths covers a wide range of topics including real numbers, polynomials, linear equations, quadratic equations, arithmetic progressions, triangles, coordinate geometry, trigonometry, circles, constructions, areas related to circles, and surface areas and volumes.

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Class 10 Maths Chapter 5 Case Based Questions - Arithmetic Progressions

Case Study - 1

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, answer the following:

Case Study - 2

Case Study - 3

Based on the above information, answer the following questions:

Q1: Find the production during first year. Ans: We are given that the production of TV sets in the factory is increasing uniformly. This means that we can apply the formula of an arithmetic progression, where the nth term (a_n) is given by a + (n-1)d. Here, 'a' is the first term, 'n' is the term number, and 'd' is the common difference. We know that the 6th year production (a_6) is 16000 and the 9th year production (a_9) is 22600. We can subtract the 6th year production from the 9th year to find the total increase over 3 years, which is 6600. This means that the common difference ('d') is 6600/3 = 2200 sets per year. Substituting the values into the nth term formula: 16000 = a + (6-1)*2200 16000 = a + 11000 => a = 16000 - 11000 => a = 5000 So, the production during the first year was 5000 sets.   Q2: Find the production during 8th year. Ans: We can use the nth term formula again to find the production in the 8th year. a_8 = a + (8-1)*d = 5000 + 7*2200 = 5000 + 15400 = 20400 So, the production during the 8th year was 20400 sets.   Q3: Find the production during first 3 years. Ans:  To find the total production over the first 3 years, we sum the production over each year. In an arithmetic progression, the sum of the first n terms (S_n) is given by n/2 * (2a + (n-1)d). S_3 = 3/2 * (2*5000 + (3-1)*2200) = 1.5 * (10000 + 4400) = 1.5 * 14400 = 21600 So, the total production during the first 3 years was 21600 sets.   Q4: In which year, the production is Rs 29,200. Ans: To find the year when the production was 29200, we can use the nth term formula and solve for n. 29200 = 5000 + (n-1)2200 => 24200 = (n-1)2200 => n-1 = 24200/2200 => n-1 = 11 => n = 12 So, the production was 29200 sets in the 12th year.   Q5: Find the difference of the production during 7th year and 4th year. Ans: We can find the production in the 7th and 4th years using the nth term formula, and then subtract the two. a_7 = 5000 + 6*2200 = 18200 a_4 = 5000 + 3*2200 = 11600 Difference = a_7 - a_4 = 18200 - 11600 = 6600 So, the difference in production between the 7th and 4th years was 6600 sets.  

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• ML Aggarwal Solutions
• ML Aggarwal Class 10 Solutions
• Chapter 9 Arithmetic and Geometric Progression

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Arithmetic and Geometric Progression

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Arithmetic and Geometric Progression are provided here. The BYJU’S team of subject-matter experts has prepared these solutions, which help students attain good marks in Maths. From the exam point of view, the solutions are solved in a simple manner so that students can secure an excellent score by solving the ML Aggarwal textbook. Solutions that are provided here will help you in getting acquainted with a wide variety of questions, and thus develop your problem-solving skills. Download the PDF of ML Aggarwal Solutions for Class 10 Maths Chapter 9 in their respective links.

Chapter 9 – Arithmetic and Geometric Progression. Arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus the first. Geometric progression is defined as a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In these ML Aggarwal Solutions for Class 10 Maths Chapter 9, students are going to learn to solve different types of problems and topics of Arithmetic and Geometric Progression.

ML Aggarwal Solutions for Class 10 Maths Chapter 9

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Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 9 Arithmetic and Geometric Progression

Exercise 9.1

1. For the following A.P.s, write the first term ‘a’ and the common difference ‘d’:

(i) 3, 1, – 1, – 3, …

From the question,

The first term a = 3

Then, difference d = 1 – 3 = – 2

– 1 – 1 = – 2

– 3 – (-1) = – 3 + 1 = -2

Therefore, common difference d = – 2

(ii) 1/3, 5/3, 9/3, 13/3, ….

The first term a = 1/3

Then, difference d = 5/3 – 1/3 = (5 – 1)/3 = 4/3

9/3 – 5/3 = (9 – 5)/3 = 4/3

13/3 – 9/3 = (13 – 9)/3 = 4/3

Therefore, common difference d = 4/3

(iii) -3.2, -3, -2.8, -2.6, ….

The first term a = -3.2

Then, difference d = -3 – (-3.2) = -3 + 3.2 = 0.2

-2.8 – (-3) = -2.8 + 3 = 0.2

-2.6 – (-2.8) = -2.6 + 2.8 = 0.2

Therefore, common difference d = 0.2

2. Write first four terms of the A.P., when the first term a and the common difference d are given as follows:

(i) a = 10, d = 10

From the question it is given that,

First term a = 10

Common difference d = 10

Then the first four terms are = 10 + 10 = 20

20 + 10 = 30

30 + 10 = 40

Therefore, first four terms are 10, 20, 30 and 40.

(ii) a = -2, d = 0

First term a = -2

Common difference d = 0

Then the first four terms are = -2 + 0 = -2

-2 + 0 = -2

Therefore, first four terms are -2, -2, -2 and -2.

(iii) a = 4, d = -3

First term a = 4

Common difference d = -3

Then the first four terms are = 4 + (-3) = 4 – 3 = 1

1 + (-3) = 1 – 3 = – 2

-2 + (-3) = -2 – 3 = – 5

Therefore, first four terms are 4, 1, -2 and -5.

(iv) a = ½, d = -1/6

First term a = ½

Common difference d = -1/6

Then the first four terms are = ½ + (-1/6) = ½ – 1/6 = (3 – 1)/6 = 2/6 = 1/3

1/3 + (-1/6) = 1/3 – 1/6 = (2 – 1)/6 = 1/6

1/6 + (-1/6) = 1/6 – 1/6 = 0

Therefore, first four terms are ½, 1/3, 1/6 and 0.

3. Which of the following lists of numbers form an A.P.? If they form an A.P., find the common difference d and write the next three terms:

(i) 4, 10, 16, 22, …

Then, difference d = 10 – 4 = 6

16 – 10 = 6

22 – 16 = 6

Therefore, common difference d = 6

Hence, the numbers are form A.P.

(ii) -2, 2, -2, 2, …

Then, difference d = -2 – 2 = – 4

-2 – 2 = -4

2 – (-2) = 2 + 2 = 4

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

(iii) 2, 4, 8, 16, …

First term a = 2

Then, difference d = 4 – 2 = 2

8 – 4 = 4

16 – 8 = 8

(iv) 2, 5/2, 3, 7/2, …

Then, difference d = 5/2 – 2 = (5 – 4)/2 = ½

3 – 5/2 = (6 – 5)/2 = ½

7/2 – 3 = (7 – 6)/2 = ½

Therefore, common difference d = ½

(v) – 10, -6, -2, 2, …

First term a = -10

Then, difference d = -6 – (- 10) = – 6 + 10 = 4

-2 – (-6) = – 2 + 6 = 4

Therefore, common difference d = 4

(vi) 1 2 , 3 2 , 5 2 , 7 2 , …

First term a = 1 2 = 1

Then, difference d = 3 2 – 1 2 = 9 – 1 = 8

5 2 – 3 2 = 25 – 9 = 16

7 2 – 5 2 = 49 – 25 = 24

Exercise 9.2

1. Find the A.P. whose n th term is 7 – 3K. Also find the 20th term.

n th term is 7 – 3k

So, T n = 7 – 3n

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

T 1 = 7 – (3 × 1) = 7 – 3 = 4

T 2 = 7 – (3 × 2) = 7 – 6 = 1

T 3 = 7 – (3 × 3) = 7 – 9 = -2

T 4 = 7 – (3 × 4) = 7 – 12 = – 5

T 20 = 7 – (3 × 20) = 7 – 60 = – 53

Therefore, A.P. is 4, 1, -2, -5, …

So, 20 th term is – 53

2. Find the indicated terms in each of following A.P.s:

(i) 1, 6, 11, 16, …; a 20

The first term a = 1

Then, difference d = 6 – 1 = 5

11 – 6 = 5

16 – 11 = 5

Therefore, common difference d = 5

From the formula, a n = a + (n – 1)d

So, a 20 = a + (20 – 1)d

= 1 + (20 – 1)5

= 1 + (19)5

Therefore, a 20 = 96

(ii) -4, -7, -10, -13, …, a 25 , a n

The first term a = -4

Then, difference d = -7 – (-4) = – 7 + 4 = -3

-10 – (-7) = -10 + 7 = -3

-13 – (-10) = -13 + 10 = -3

Therefore, common difference d = -3

So, a 25 = a + (25 – 1)d

= -4 + (25 – 1)(-3)

= -4 + (24)-3

= – 4 – 72

Therefore, a 25 = -76

Now, a n = a + (n – 1)d

a n = -4 + (n – 1)-3

= -4 – 3n + 3

= -1 – 3n

3. Find the n th term and the 12 th term of the list of numbers: 5, 2, – 1, – 4, …

The first term a = 5

Then, difference d = 2 – 5 = – 3

-1 – 3 = -3

– 4 – (-1) = -4 + 1 = -3

T n = a + (n – 1)d

= 5 + (n – 1)-3

= 5 – 3n + 3

= 8 – 3n

So, T 12 = a + (12 – 1)d

= 5 + (12 – 1)(-3)

= 5 + (11)-3

= 5 – 33

= – 28

4. (i) If the common difference of an A.P. is – 3 and the 18 th term is – 5, then find its first term.

The 18 th term = -5

Then, common difference d = -3

So, T 18 = a + (18 – 1)d

-5 = a + (18 – 1)(-3)

-5 = a + (17)(-3)

-5 = a – 51

a = 51 – 5

Therefore, first term a = 46

(ii) If the first term of an A.P. is – 18 and its 10th term is zero, then find its common difference.

The 10 th term = 0

Then, first term a = -18

So, T 10 = a + (10 – 1)d

0 = -18 + (10 – 1)d

0 = -18 + 9d

Therefore, common difference d = 2

5. Which term of the A.P.

(i) 3, 8, 13, 18, … is 78?

Let us assume 78 as n th term.

Then, difference d = 8 – 3 = 5

13 – 8 = 5

18 – 13 = 5

So, 78 = a + (n – 1)d

78 = 3 + (n – 1)5

78 = 3 + 5n – 5

78= -2 + 5n

5n = 78 + 2

Therefore, 78 is 16 th term.

(ii) 18, 15½, 13, … is – 47?

Convert mixed fraction into improper fraction = 15½ = 31/2

Let us assume -47 as n th term.

The first term a = 18

Then, difference d = 31/2 – 18 = (31 – 36)/2 = -5/2

13 – 31/2 = (26 – 31)/2 = -5/2

Therefore, common difference d = -5/2

So, -47 = a + (n – 1)d

-47 = 18 + (n – 1) (-5/2)

-47 = 18 – 5/2n + 5/2

-47 – 18 = -5/2n + 5/2

-65 = -5/2n + 5/2

-65 – 5/2 = – 5/2n

(-130 – 5)/2 = -5/2n

-135/2 = -5/2n

n = (-135/2) × (-2/5)

n = -135/-5

Therefore, -47 is 27 th term.

6. (i) Check whether – 150 is a term of the A.P. 11, 8, 5, 2, …

The first term a = 11

Then, difference d = 8 – 11 = -3

5 – 8 = -3

2 – 5 = -3

Then, common difference d = – 3

Let us assume -150 as n th term,

So, -150 = 11 + (n – 1) (-3)

-150 = 11 – 3n + 3

-150 = 14 – 3n

3n = 150 + 14

Therefore, – 150 is not a term of the A.P. 11, 8, 5, 2, …

(ii) Find whether 55 is a term of the A.P. 7, 10, 13, … or not. If yes, find which term is it.

The first term a = 7

Then, difference d = 10 – 7 = 3

13 – 10 = 3

Then, common difference d = 3

Let us assume 55 as n th term,

So, 55 = 7 + (n – 1)3

55 = 7 + 3n – 3

55 = 4 + 3n

3n = 55 – 4

Therefore, 55 is 17 th term of the A.P. 7, 10, 13, …

7. (i) Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.

Let us assume 253 as n th term.

So, 253 = a + (n – 1)d

253 = 3 + (n – 1)5

253 = 3 + 5n – 5

253 = -2 + 5n

5n = 253 + 2

Therefore, 253 is 51 th term.

Now, assume ‘P’ be the 20 th term from the last.

Then, P = L – (n – 1)d

= 253 – (20 – 1) 5

= 253 – (19) 5

= 253 – 95

Therefore, 158 is the 20 th term from the last.

(ii) Find the 12 th from the end of the A.P. – 2, – 4, – 6, …, – 100.

Let us assume -100 as n th term.

The first term a = -2

Then, difference d = – 4 – (-2) = – 4 + 2 = -2

-6 – (-4) = -6 + 4 = – 2

Therefore, common difference d = -2

So, – 100 = a + (n – 1)d

– 100 = -2 + (n – 1)(-2)

– 100 = -2 – 2n + 2

– 100 = -2n

n = -100/-2

Therefore, -100 is 50 th term.

Now, assume ‘P’ be the 12 th term from the last.

= -100 – (12 – 1) (-2)

= -100 – (11) (-2)

= -100 + 22

P = – 78

Therefore, -78 is the 12 th term from the last of the A.P. – 2, – 4, – 6, …

8. Find the sum of the two middle most terms of the A.P.

First term a = -4/3

Then, difference d = -1 – (-4/3) = -1 + 4/3 = (-3 + 4)/3 = 1/3

= -2/3 – (-1) = -2/3 + 1 = (-2 + 3)/3 = 1/3

Therefore, common difference d = 1/3

We know that,

So, 13/3 = -4/3 + (n – 1)(1/3)

13/3 + 4/3 = 1/3n – 1/3

13/3 + 4/3 + 1/3 = 1/3n

(13 + 4 + 1)/3 = 1/3n

18/3 = 1/3n

n = 6 × 3

So, middle term is 18/2 and (18/2) + 1 = 9 th and 10 th term

Then, a 9 + a 10 = a + 8d + a + 9d

Now substitute the value of ‘a’ and ‘d’.

= 2(-4/3) + 17(1/3)

= -8/3 + 17/3

= (-8 + 17)/3

Therefore, the sum of the two middle most terms of the A.P is 3.

9. Which term of the A.P. 53, 48, 43,… is the first negative term ?

The first term a = 53

Then, difference d = 48 – 53 = -5

= 43 – 48 = -5

Therefore, common difference d = -5

= 53 + (n – 1)(-5)

= 53 – 5n + 5

= 58 – 5n

n = 11.6 ≈ 12

Therefore, 12 th term is the first negative term of the A.P. 53, 48, 43,…

10. Determine the A.P. whose third term is 16 and the 7 th term exceeds the 5 th term by 12.

The 7th term exceeds the 5th term by 12 = T 7 – T 5 = 12

We know that, T n = a + (n – 1)d

So, T 3 = a + 2d = 16 … [equation (i)]

T 7 – T 5 = (a + 6d) – (a + 4d) = 12 … [equation (ii)]

12 = a + 6d – a – 4d

Now, substitute value of d in equation (i) we get,

Then, T 3 = a + 2d

16 = a + 2(6)

a = 16 – 12

Therefore, A.P. is 4 + 6 = 10, 10 + 6 = 16, 16 + 6 = 22

Hence, the four term of A.P. is 4, 10, 16, 22, ….

11. Find the 20 th term of the A.P. whose 7 th term is 24 less than the 11 th term, first term being 12.

First term a = 12

7th term is 24 less than the 11th term = T 11 – T 7 = 24

T 11 – T 7 = (a + 10d) – (a + 6d) = 24

24 = a + 10d – a – 6d

Now, T 20 = a + 19d

Substitute the values of a and d,

T 20 = 12 + 19(6)

T 20 = 12 + 114

12. Find the 31 st term of an A.P. whose 11 th term is 38 and 6 th term is 73.

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So, T 11 = a + 10d = 38 equation (i)

T 6 = a + 5d = 73 equation (ii)

Subtracting both equation (i) and equation (i),

(a + 10d) – (a + 5d) = 73 – 38

a + 10d – a – 5d = 35

now, substitute the value of d in equation (i) to find out a, we get

a + 10d = 38

a + 10(7) = 38

a + 70 = 38

a = 38 – 70

Therefore, T 31 = a + 30d

= -32 + 30(7)

= -32 + 210

13. If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its 63 rd term.

So, T 9 = a + 8d = 1/7 equation (i)

T 7 = a + 6d = 1/9 equation (ii)

(a + 6d) – (a + 8d) = 1/9 – 1/7

a + 6d – a – 8d = (7 – 9)/63

-2d = -2/63

d = (-2/63) × (-1/2)

now, substitute the value of d in equation (ii) to find out a, we get

a + 6(1/63) = 1/9

a = 1/9 – 6/63

a = (7 – 6)/63

Therefore, T 63 = a + 62d

= 1/63 + 62(1/63)

= 1/63 + 62/63

= (1 + 62)/63

14. (i) The 15 th term of an A.P. is 3 more than twice its 7 th term. If the 10 th term of the A.P. is 41, find its n th term.

From the question it I s given that,

T 10 = a + 9d = 41 … [equation (i)]

T 15 = a + 14d = 2T 7 + 3

= a + 14d = 2(a + 6d) + 3

= a + 14d = 2a + 12 d + 3

-3 = 2a – a + 12d – 14d

a – 2d = -3 … [equation (ii)]

Now, subtracting equation (ii) from (i), we get,

(a + 9d) – (a – 2d) = 41 – (-3)

a + 9d – a + 2d = 41 + 3

Then, substitute the value of d is equation (i) to find a,

a + 9(4) = 41

a + 36 = 41

a = 41 – 36

Therefore, n th term = T n = a + (n – 1)d

= 5 + (n – 1)4

= 5 + 4n – 4

(ii) The sum of 5 th and 7 th terms of an A.P. is 52 and the 10 th term is 46. Find the A.P.

a 5 + a 7 = 52

(a + 4d) + (a + 6d) = 52

a + 4d + a + 6d = 52

2a + 10d = 52

Divide both the side by 2 we get,

a + 5d = 26 … equation (i)

Given, a 10 = a + 9d = 46

a + 9d = 46 … equation (ii)

Now subtracting equation (i) from equation (ii),

(a + 9d) – (a + 5d) = 46 – 26

a + 9d – a – 5d = 20

Substitute the value of d in equation (i) to find out a,

a + 5d = 26

a + 5(5) = 26

a + 25 = 26

a = 26 – 25

Then, a 2 = a + d

= 1 + 5 = 6

a 3 = a 2 + d

a 4 = a 3 + d

Therefore, 1, 6, 11, 16,… are A.P.

15. If 8 th term of an A.P. is zero, prove that its 38 th term is triple of its 18 th term.

We have to prove that, 38 th term is triple of its 18 th term = T 38 = 3T 18

T 8 = a + 7d = 0

T 8 = a = -7d

T 38 = a + 37d

= -7d + 37d

Take, T 18 = a + 17d

Substitute the value of a and d,

T 18 = -7d + 17d

By comparing the results of T 38 and T 18 , the 38 th term is triple its 18 th term.

16. Which term of the A.P. 3, 10, 17,… will be 84 more than its 13 th term?

From the question, it is given that,

First term a = 3

Common difference d = 10 – 3 = 7

Then, T 13 = a + 12d

= 3 + 12(7)

Let us assume that n th term is 84 more than its 13 th term

So, T n = 84 + 87

We know that, T n = a + (n – 1)d = 171

3 + (n – 1)7 = 171

3 + 7n – 7 = 171

7n – 4 = 171

7n = 171 + 4

17. (i) How many two digit numbers are divisible by 3 ?

The two digits numbers divisible by 3 are, 12, 15, 18, 21, 24,…..,99.

The above numbers are A.P.

So, first number a = 12

Common difference d = 15 – 12 = 3

Then, last number is 99

We know that, T n (last number) = a + (n – 1)d

99 = 12 + (n – 1)3

99 = 12 + 3n – 3

99 = 9 + 3n

99 – 9 = 3n

Therefore, 30 two digits number are divisible by 3.

(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

The natural numbers which are divisible by both 2 and 5 are 110, 120, 130, 140, ….,999

So, first number a = 110

Common difference d = 120 – 110 = 10

Then, last number is 999

999 = 110 + (n – 1)10

999 = 110 + 10n – 10

999 = 100 + 10n

999 – 100 = 10n

The number of natural numbers which are divisible by both 2 and 5 are 88.

18. If the numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n.

From the question it is given that, n – 2, 4n – 1 and 5n + 2 are in A.P.

Multiplying by 2 to 4n – 1 then it becomes = 8n – 2

So, 8n – 2 = n – 2 + 5n + 2

8n – 2 = 6n

8n – 6n = 2

19. The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.

The sum of three numbers in A.P. = 3

Given, Their product = -35

Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d

Now adding 3 numbers = a – d + a + a + d = 3

From the question, product of 3 numbers is – 35

So, (a – d) × (a) × (a + d) = – 35

(1 – d) × (1) × (1 + d) = – 35

1 2 – d 2 = – 35

d 2 = 35 + 1

d = √36

d = ±6

Therefore, the numbers are (a – d) = 1 – 6 = – 5

(a + d) = 1 + 6 = 7

If d = – 6

The numbers are (a – d) = 1 – (-6) = 1 + 6 = 7

(a + d) = 1 + (-6) = 1 – 6 = -5

Therefore, the numbers -5, 1, 7,… and 7, 1, -5,… are in A.P.

20. The sum of three numbers in A.P. is 30 and the ratio of first number to the third number is 3 : 7. Find the numbers.

From the question it is given that, sum of three numbers in A.P. = 30

The ratio of first number to the third number is 3: 7

Now adding 3 numbers = a – d + a + a + d = 30

Given ratio 3 : 7 = a – d : a + d

3/7 = (a – d)/(a + d)

(a – d)7 = 3(a + d)

7a – 7d = 3a + 3d

7a – 3a = 7d + 3d

4(10) = 10d

Therefore, the numbers are a – d = 10 – 4 = 6

a + d = 10 + 4 = 14

So, 6, 10, 14, … are in A.P.

21. The sum of the first three terms of an A.P.is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.

From the question it is given that sum of the first three terms of an A.P. is 33.

Now adding 3 numbers = a – d + a + a + d = 33

Given, the product of the first and the third terms exceeds the second term by 29.

(a – d) (a + d) = a + 29

a 2 – d 2 = 11 + 29

11 2 – d 2 = 40

121 – 40 = d 2

d =√81

d = ±9

Therefore, the numbers are (a – d) = 11 – 9 = 2

(a + d) = 11 + 9 = 20

If d = – 9

The numbers are (a – d) = 1 – (-9) = 11 + 9 = 20

(a + d) = 11 + (-9) = 11 – 9 = 2

Therefore, the numbers 2, 11, 20,… and 20, 11, 2,… are in A.P.

Exercise 9.3

1. Find the sum of the following A.P.s : (i) 2, 7, 12, … to 10 terms

Then, d = 7 – 2 = 5

12 – 7 = 5

So, common difference d = 5

S 10 = n/2(2a + (n – 1)d)

= 10/2 ((2 × 2) + (10 – 1)5)

= 5(4 + 45)

(ii) 1/15, 1/12, 1/10, … to 11 terms

First term a = 1/15

Then, d = 1/12 – 1/15

= (5 – 4)/60

So, common difference d = 1/60

S 11 = 11/2(2a + (n – 1)d)

= 11/2 ((2 × (1/15)) + (11 – 1)(1/60))

= 11/2 ((2/15) + (10/60))

= 11/2 (2/15 + 1/6)

= 11/2 (4 + 5)/30

= 11/2 (9/30)

= 11/2(3/10)

2. Find the sums given below : (i) 34 + 32 + 30 + … + 10

First term a = 34,

Difference d = 32 – 34 = -2

So, common difference d = – 2

Last term T n = 10

10 = 34 + (n – 1)(-2)

-24 = – 2(n – 1)

-24 = – 2n + 2

2n = 24 + 2

S n = n/2(a + 1)

= 13/2 (34 + 10)

= 13/2 (44)

(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)

First term a = -5,

Difference d = -8 – (-5) = -8 + 5 = -3

So, common difference d = – 3

Last term T n = -230

-230 = -5 + (n – 1)(-3)

-230 = – 5 – 3n + 3

-230 = – 2 – 3n

3n = 230 -2

Therefore, S n = n/2 (a + l)

= 76/2 (-5 + (-230))

= 38 (-5 – 230)

= – 8930

3. In an A.P. (with usual notations) : (i) given a = 5, d = 3, a n  = 50, find n and S n

First term a = 5

Then common difference d = 3

We know that, a n = a + (n – 1)d

50 = 5 + (n – 1)3

50 = 5 + 3n – 3

50 = 2 + 3n

3n = 50 – 2

So, S n = (n/2)(2a + (n – 1)d)

= (16/2) ((2 × 5) + (16 – 1) × 3)

= 8(10 + 45)

(ii) given a = 7, a 13  = 35, find d and S 13

First term a = 7

35 = 7 + (13 – 1)d

35 = 7 + 13d – d

35 = 7 + 12d

12d = 35 – 7

d = 28/12 … [divide by 4]

So, S 13 = (n/2)(2a + (n – 1)d)

= (13/2) ((2 × 7) + ((13 – 1) × (7/3))

= (13/2) ((14 + (12 × 7/3))

= (13/2) (14 + 28)

= (13/2) (42)

= 13 × 21

(iii) given d = 5, S 9  = 75, find a and a 9 .

Common difference d = 5

a 9 = a + (9 – 1)5

a 9 = a + 45 – 5

a 9 = a + 40 … [equation (i)]

Then, S 9 = (n/2) (2a + (n – 1)d)

75 = (9/2) (2a + (9 – 1)5)

75 = (9/2) (2a + (8)5)

(75 × 2)/9 = 2a + 40

150/9 = 2a + 40

2a = 150/9 – 40

2a = 50/3 – 40

2a = (50 – 120)/3

a = -70/(3 × 2)

a = – 35/3

Now, substitute the value of a in equation (i),

a 9 =a + 40

= -35/3 + 40

= (-35 + 120)/3

(iv) given a = 8, a n  = 62, S n  = 210, find n and d

Solution:- From the question it is give that,

First term a = 8,

a n = 62 and S n = 210

62 = 8 + (n – 1)d

(n – 1)d = 62 – 8

(n – 1)d = 54 … [equation (i)]

Then, S n = (n/2) (2a + (n – 1)d)

210 = (n/2) ((2 × 8) + 54) … [from equation (i) (n – 1)d = 54]

210 = (n/2) (16 + 54)

420 = n(70)

Now, substitute the value of n in equation (i),

(n – 1)d = 54

(6 – 1)d = 54

Therefore, d = 54/5 and n = 6

(v) given a = 3, n = 8, S = 192, find d.

We know that, S n = (n/2) (2a + (n – 1)d)

192 = (8/2) ((2 × 3) + (8 – 1)d)

192 = 4 (6 + 7d)

192/4 = 6 + 7d

48 = 6 + 7d

48 – 6 = 7d

Therefore, common difference d is 6.

4. (i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

From the question it is give that,

Last term = 45

Then, sum = 400

We know that, last term = a + (n – 1)d

45 = 5 + (n – 1)d

(n – 1)d = 45 – 5

(n – 1)d = 40 … [equation (i)]

So, S n = (n/2) (2a + (n – 1)d)

400 = (n/2) ((2 × 5) + 40) … [from equation (i) (n – 1)d = 40]

800 = n(10 + 40)

(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

First term a = 15

Therefore, the sum of the first n terms of an A.P. is given by,

S n = (n/2) (2a + (n – 1)d)

S 15 = (15/2)(2a + (15 – 1)d)

750 = (15/2) (2a + 14d)

(750 × 2)/15 = 2a + 14d

100 = 2a + 14d

Dividing both the side by 2 we get,

50 = a + 7d

Now, substitute the value a,

50 = 15 + 7d

7d = 50 – 15

So, 20 th term a 20 = a + 19d

= 15 + 19(5)

5. The first and the last terms of an A.P. are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?

First term a = 17

Last term (l) = 350

Common difference d = 9

We know that, l = T n = a + (n – 1)d

350 = 17 + (n – 1) × 9

350 – 17 = 9n – 9

333 + 9 = 9n

= (38/2) ((2 × 17) + (38 – 1)d)

= 19(34 + (37 × 9))

= 19(34 + 333)

= 19 × 367

Therefore, n = 38 and S n = 6973

6. Solve for x : 1 + 4 + 7 + 10 + … + x = 287.

First term a = 1

Difference d = 4 – 1 = 3

x = a = (n – 1)d

x – 1 = (n – 1)d

287 = (n/2) ((2 × 1)+ (n – 1)3)

= n (2 + 3n – 3)

574 = n(2 + 3n – 3)

574 = 2n + 3n 2 – 3n

574 = – n + 3n 2

3n 2 – n – 574 = 0

3n 2 – 42n + 41 – 574 = 0

3n(n – 14) + 41(n – 14) = 0

(n – 14) (3n + 41) = 0

If n – 14 = 0

or 3n + 41 = 0

We have to take a positive number, so n = 14

Then, = a + (n – 1)d

= 1 + (14 – 1) 3

= 1 + (13)3

Therefore, x = 40

7. (i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.

First term a = 25

Common difference d = 22 – 25 = – 3

116 = (n/2) (2a + (n – 1)d)

By cross multiplication,

232 = n ((2 × 25) + (n – 1) (-3))

232 = n (50 – 3n + 3)

232 = n (53 – 3n)

232 = 53n – 3n 2

3n 2 – 53n + 232 = 0

3n 2 – 24n – 29n + 232 = 0

3n (n – 8) – 29 (n – 8) = 0

(n – 8) (3n – 29) = 0

If n – 8 = 0

or 3n – 29 = 0

not possible to take fraction,

Then, T = a + (n – 1)d

= 25 +(8 – 1) (-3)

= 25 + 7 (-3)

= 25 – 21

(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.

First term a = 24

Common difference d = 21 – 24 = – 3

78 = (n/2) (2a + (n – 1)d)

156 = n ((2 × 24) + (n – 1) (-3))

156 = n (48 – 3n + 3)

156 = n (51 – 3n)

156 = 51n – 3n 2

3n 2 – 51n + 156 = 0

3n 2 – 12n – 39n + 156 = 0

3n (n – 4) – 39 (n – 4) = 0

(n – 4) (3n – 39) = 0

If n – 4 = 0

or 3n – 39 = 0

now we have to consider both values

= 24 +(4 – 1) (-3)

= 24 + 3 (-3)

= 24 – 9

= 24 +(13 – 1) (-3)

= 24 + 12 (-3)

= 24 – 36

So, (12 + 9 + 6 + 3 + 0 + (-3)+ (-6) + (-9) + (-12)) = 0

Hence, the sum of 5 th term to 13 th term = 0

8. Find the sum of first 22 terms, of an A.P. in which d = 7 and a 22  is 149.

Common difference d = 7

we know that,

a 22 = (n – 1)d

149 = a + (22 – 1)7

149 = a + (22)7

149 = a + 147

a = 149 – 147

So, S 22 = (n/2) (2a + (n – 1)d)

= (22/2) ((2 × 2) + (22 – 1)7)

= 11(4 + (21)7)

= 11 (4 + 147)

9. In an A.P., the fourth and sixth terms are 8 and 14, respectively. Find the:

(i) first term (ii) common difference

(iii) sum of the first 20 terms.

T 4 = 8 and T 6 = 14

⇒ a + 3d = 8 … (i)

⇒ a + 5d = 14 … (ii)

Subtracting (i) from (ii), we get

(a + 5d) – (a + 3d) = 14 – 8

5d – 3d = 6

(ii) Hence, common difference d = 3

Substituting the value of d in (i), we have

a + 3(3) = 8

a = 8 – 9

(i) Hence, the first term = -1

(iii) The sum of first 20 terms

S 20 = n/2 × [2a + (n – 1)d]

= 20/2 × [2(-1) + (20 – 1)(3)]

= 10 × [-2 + (19)(3)]

= 10 × [-2 + 57]

= 10 × 55

10. (i) Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18, respectively.

T 2 = 14, T 3 = 18

So, common difference d = T 3 – T 2

= 18 – 14

Where, a = T 1 = 14 – 4 = 10

S 51 = (n/2) (2a + (n – 1)d)

= (51/2) ((2 × 10) + (51 – 1)4)

= (51/2) (20 + (50 × 4))

= (51/2) (20 + 200)

= (51/2) × 220

(ii) The 4 th term of A.P is 22 and 15 th term is 66. Find the first term and the common difference. Hence, find the sum of first 8 term of the A.P.

T 4 = 22, T 15 = 66

⇒ a + 3d = 22 … (i)

⇒ a + 14d = 66 … (ii)

(a + 14d) – (a + 3d) = 66 – 22

14d – 3d = 44

So, common difference d = 4

a + 3(4) = 22

a = 22 – 12

Hence, the first term a = 10

Now, the sum of first 8 terms of the A.P is

S n = n/2 × [2a + (n – 1)d]

S 8 = 8/2 × [2(10) + (8 – 1)(4)]

= 10 × [20 + (7)(4)]

= 10 × [20 + 28]

= 10 × 48

11. If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

S 6 = (6/2) (2a + (6 – 1)d) = 36

3 (2a + 5d) = 36

Divide both the side by 3,

2a + 5d = 12 … [equation (i)]

Now, S 16 = (16/2) (2a + (16 – 1)d) = 256

8 (2a + 15d) = 256

Divide both the side by 8,

2a + 15d = 32 … [equation (ii)]

Then, subtract equation (ii) from equation (i) we get,

(2a + 5d) – (2a + 15d) = 12 – 32

2a + 5d – 2a – 15d = -20

d = -20/-10

substitute the value of d in equation (i) to find a,

2a + 5d = 12

2a + 5(2) = 12

2a + 10 = 12

2a = 12 – 10

So, S 10 = (n/2) (2a + (n – 1)d)

= (10/2) ((2 × 1) + (10 – 1)2)

= 5 (2 + 18)

Therefore, the sum of first 10 terms is 100.

12. Show that a 1 , a 2 , a 3 , … form an A.P. where a n  is defined as a n  = 3 + 4n. Also find the sum of first 15 terms.

n th term is 3 + 4n

So, a n = 3 + 4n

a 1 = 3 + (4 × 1) = 3 + 4 = 7

a 2 = 3 + (4 × 2) = 3 + 8 = 11

a 3 = 3 + (4 × 3) = 3 + 12 = 15

a 4 = 3 + (4 × 4) = 3 + 16 = 19

So, The numbers are 7, 11, 15, 19, ….

Then, first term a = 7, common difference d = 11 – 7 = 4

S 15 = (n/2) (2a + (n – 1)d)

= (15/2) ((2 × 7) + (15 – 1) × 4)

= (15/2) (14 + (14 × 4))

= (15/2) (14 + 56)

= (15/2) × 70

Therefore, the sum of first 15 terms is 525.

13. The sum of first six terms of an arithmetic progression is 42. The ratio of the 10 th term to the 30 th term is 1: 3. Calculate the first and the thirteenth term.

S 6 = 42 and T 10 /T 30 = 1/3

So, we have

S 6 = (6/2) × [2a + (6 – 1)d] = 42

3 × (2a + 5d) = 42

2a + 5d = 14 … (i) [Dividing by 3]

T 10 /T 30 = [a + (10 – 1)d]/ [a + (30 – 1)d] = 1/3

(a + 9d)/ (a + 29d) = 1/3

On cross-multiplication, we have

3(a + 9d) = a + 29d

3a + 27d = a + 29d

3a – a + 27d – 29d = 0

2a – 2d = 0

a – d = 0

Using the above the relation in (i), we get

2a + 5a = 14

Hence, the first term of the A.P is 2.

Now, the thirteenth term is given by

T 13 = 2 + (13 – 1)(2)

= 2 + 12 × 2

14. In an A.P., the sum of its first n terms is 6n – n 2 . Find its 25 th term.

S n = 6n – n 2

S 1 = 6(1) – (1) 2 = 6 – 1 = 5

So, the first term a = 5

S 2 = 6(2) – (2) 2 = 12 – 4 = 8

So, T 1 + T 2 = 8

5 + T 2 = 8

T 2 = 8 – 5 = 3

So, the common difference d = T 2 – T 1

d = 3 – 5

Thus, 25 th term in the A.P is given by

T 25 = 5 + (25 – 1)(-2)

= 5 + 24(-2)

= 5 – 48

15. If S n denotes the sum of first n terms of an A.P., prove that S 30 = 3 (S 20 – S 10 ).

S 10 = 10/2 × [2a + (10 – 1)d]

= 5 × (2a + 9d)

= 10a + 45d

S 20 = 20/2 × [2a + (20 – 1)d]

= 10 × (2a + 19d)

= 20a + 190d

S 30 = 30/2 × [2a + (30 – 1)d]

= 15 × (2a + 29d)

= 30a + 435d

Now, taking L.H.S we have

3 (S 20 – S 10 ) = 3 [20a + 190d – (10a + 45d)]

= 3 (20a + 190d – 10a – 45d)

= 3 (10a + 145d)

– Hence proved

16. (i) Find the sum of first 1000 positive integers.

(ii) Find the sum of first 15 multiples of 8.

(i) First 1000 positive integers are:

1, 2, 3, 4, …….., 1000

This is an A.P with first term a = 1 and common difference d = 1

S 1000 = 1000/2 × [2(1) + (1000 – 1)(1)]

= 500 × (2 + 99)

= 500 × 101

Therefore, the sum of first 1000 positive integers is 50500.

(ii) The first 15 multiples of 8 are:

8, 16, 24, ….. where n = 15

This forms an A.P with first term a = 8 and common difference d = 16 – 8 = 8

S 15 = 15/2 × [2(8) + (14 – 1)(8)]

= 15/2 × (16 + 13 × 8)

= 15/2 × (16 + 104)

= 15/2 × 120

= 15 × 60

Therefore, the sum of first 15 multiples of 8 is 900.

17. (i) Find the sum of all two digit natural numbers which are divisible by 4.

(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

(iii) Find the sum of all multiples of 9 lying between 300 and 700.

(iv) Find the sum of all natural numbers less than 100 which are divisible by 4.

(i) The two-digit natural numbers which are divisible by 4 are:

4, 8, 12, 16, …..

This form an A.P.

The last term in this series is found out by dividing 100 by 4

100/4 = 25 and the remainder is zero

So, the last two-digit number which is divisible by 4 is 96

And, it is the 24 th term

a = 4, d = 4, last term l = 96 and n = 24

Thus, sum of these numbers is

S 24 = 24/2 (4 + 96)

= 12 × 100

(ii) The natural numbers between 100 and 200 which are divisible by 4 are:

104, 108, 112, …..

The last term in this series is found out by dividing 200 by 4

200/4 = 50 and remainder is zero

So, the last natural number between 100 and 200 which is divisible by 4 is 196

a = 104, d = 4 and last term l = 96

The number of terms is found out by

T n = 196 = 104 + (n – 1)(4)

196 = 104 + 4n – 4

4n = 196 – 104 + 4

S 24 = 24/2 (104 + 196)

= 12 × 200

(iii) The multiples of 9 lying between 300 and 700 are:

306, 315, 324, ……

This form an A.P. where a = 306 and d = 9

The last term in this series is found out by dividing 700 by 9

700/9 gives 77 as quotient and 7 as remainder.

So, the last number between 300 and 700 which is a multiple is 700 – 7 = 693

Now, we have

a = 306, d = 9 and l = 693

T n = 693 = 306 + (n – 1)(9)

693 = 306 + 9n – 9

9n = 693 – 306 + 9

S 44 = 44/2 (306 + 693)

= 22 × 999

(iv) The natural numbers less than 100 which are divisible by 4 are:

4, 8, 12, …., 96

This forms an A.P. where a = 4, d = 4 and l = 96

Now, for calculating n

T n = 96 = 4 + (n – 1)4

96 = 4 + 4n – 4

Exercise 9.4

1. (i) Find the next term of the list of numbers 1/6, 1/3, 2/3, …

First term a = 1/6

Then, r = (1/3) ÷ (1/6)

r = (1/3) × (6/1)

Therefore, next term = 2/3 × 2 = 4/3

(ii) Find the next term of the list of numbers 3/16, -3/8, ¾, -3/2,…

First term a = 3/16

Then, r = (-3/8) ÷ (3/16)

r = (-3/8) × (16/3)

r = (-3 × 16)/(8 × 3)

r = (-1 × 2)/ (1 × 1)

Therefore, next term = -3/2 × (-2) = 6/2 = 3

(iii) Find the 15 th term of the series √3 + 1/√3 + 1/3√3 + …

First term a = √3

Then, r = (1/√3) ÷ (√3)

r = (1/√3) × (1/√3)

r = (1 × 1)/( √3 × √3)

r = 1/(√3) 2

So, a 15 = ar n – 1

= √3(1/3) 15 – 1

= √3(1/3) 14

= √3 × (1/3 14 )

Therefore, a 15 = √3 × (1/3 14 )

(iv) Find the n th term of the list of numbers 1/√2, -2, 4√2, – 16,…

First term a = 1/√2

Then, r = -2 ÷ (1/√2)

r = (-2/1) × (√2/1)

r = (-2 × √2)/(1 × 1)

r = -2√2

So, a n = ar n – 1

= (1/√2)(-2√2) n – 1

= (1/√2) × (-1) n – 1 × [(√2) 2 × √2] n – 1

= (-1) n – 1 × 1/√2 × [(√2) 3 ] n – 1

= (-1) n – 1 × 1/√2 × (√2) 3n – 3

= (-1) n – 1 (√2) 3n – 3 – 1

= (-1) n – 1 (√2) 3n – 4

= (-1) n – 1 × 2 (3n – 4)/2

Therefore, a n = (-1) n – 1 × 2 (3n – 4)/2

(v) Find the 10 th and n th terms of the list of numbers 5, 25, 125, …

First term a = 5,

Then, r = (25) ÷ (5)

r = (25) × (1/5)

So, a 10 = ar n – 1

= 5 × (5) 10 – 1

= 5 × 5 9

= 5 9 + 1 … [by a m × a n = a m + n ]

Therefore, a n = ar n – 1

= 5 × 5 n – 1

= 5 n – 1 + 1

(vi) Find the 6 th and the n th terms of the list of numbers 3/2, ¾, 3/8,…

First term a = 3/2,

Then, r = (3/4) ÷ (3/2)

r = (3/4) × (2/3)

r = (3 × 2)/(4 × 3)

r = (1 × 1)/(2 × 1)

r = ½

= (3/2) × (1/2) n – 1

= 3 × ½ × (½) n – 1

= 3 × (½) n – 1 + 1

= 3 × (½) n

Therefore, a 6 = 3/2 n

(vii) Find the 6th term from the end of the list of numbers 3, – 6, 12, – 24, …, 12288.

Last term = 12288

First term a = 3,

Then, r = (-6) ÷ (3)

r = (-6) × (1/3)

r = (-6 × 1)/(1 × 3)

r = (-2 × 1)/(1 × 1)

Then, 6 th term from the end,

a 6 = l × (1/r) n – 1

= 12288 × (1/-2) 6 – 1

= 12288 × (1/-2 5 )

= 12288/-32

= – 384

2. Which term of the G.P.

(i) 2, 2√2, 4, … is 128?

Last term = 128

First term a = 2,

Then, r = (2√2) ÷ (2)

r = (2√2)/2

r = √2

Then, a n = ar n – 1

So, 128 = 2(√2) n – 1

2 7 = 2(√2) n – 1

2 7 /2 = (√2) n – 1

2 7 – 1 = (√2) n – 1

2 6 = (√2) n – 1

(√2) n -1 = (√2) 12

Now, comparing the powers

n – 1 = 12

Therefore, 128 is the 13 th term.

(ii) 1, 1/3, 1/9, … is 1/243

Last term (a n ) = 1/243

First term a = 1,

Then, r = (1/3) ÷ (1)

r = (1/3) × (1/1)

1/243 = 1 × (1/3) n – 1

(1/3) 5 = (1/3) n – 1

By comparing both left hand side and right hand side,

5 = n – 1

Therefore, 1/243 is 6 th term.

3. Determine the 12 th term of a G.P. whose 8 th term is 192 and common ratio is 2.

a 8 = 192 and r = 2

Then, by the formula a n = ar n – 1

a 8 = ar 8 – 1

192 = a(2) 8 – 1

192 = a(2) 7

a = 192/2 7

a = 192/128

Now, a 12 = (3/2)(2) 12 – 1

= (3/2) × (2) 11

= (3/2) × 2048

4. In a GP., the third term is 24 and 6 th term is 192. Find the 10 th term.

a 6 = ar 6 – 1

192 = ar 6 – 1

192 = ar 5 … [equation (i)]

Now, a 3 = ar n – 1

24 = ar 3 – 1

24 = ar 2 … [equation (ii)]

By dividing equation (i) by equation (ii)

ar 5 /ar 2 = 192/24

r 5 – 2 = 8

Now, substitute the value r in equation (i),

192 = a (2) 5

So, a 10 = ar 10 – 1

5. Find the number of terms of a G.P. whose first term is ¾, common ratio is 2 and the last term is 384.

First term of G.P. a = ¾

Common ratio (r) = 2

Last term = 384

384 = (3/4) (2) n – 1

(384 × 4)/3 = (2) n – 1

(1536)/3 = (2) n – 1

  512 = 2 n – 1

2 9 = 2 n – 1

9 = n – 1

The number of terms of a G.P. is 10.

6. Find the value of x such that,

(i) -2/7, x, -7/2 are three consecutive terms of a G.P.

x 2 = -2/7 × -7/2

x = ± 1

Therefore, x = 1 or x = – 1

(ii) x + 9, x – 6 and 4 are three consecutive terms of a G.P.

(x – 6) 2 = (x + 9) × 4

x 2 – 12x + 36 = 4x + 36

x 2 – 12x – 4x + 36 – 36 = 0

x 2 – 16x = 0

x(x – 16) = 0

Either let us take x – 16 = 16

So, x = 0, 16

(iii) x, x + 3, x + 9 are first three terms of a G.P. Find the value of x.

(x + 3) 2 = x(x + 9)

x 2 + 6x + 9 = x 2 + 9x

9 = 9x – 6x

7. If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

Now we have to prove that, x, y, z are in G.P.

a 4 = ar 4 – 1

So, a 7 = a 7 – 1

a 10 = a 10 -1

x, y, z are in G.P. then,

Then, xz = ar 3 × ar 9

= a 2 r 3 + 9

y 2 = (ar 6 ) 2

y 2 = a 2 r 12

By comparing left hand side and right hand side

Therefore, x, y and z are in G.P.

8. The 5 th , 8 th and 11 th terms of a G.P. are p, q and s, respectively. Show that q² = ps.

Now we have to prove that, q² = ps

a 5 = ar 5 – 1

So, a 8 = a 8 – 1

a 11 = a 11 -1

a 11 = a 10

p, q, s are in G.P. then,

q 2 = (ar 7 ) 2

Then, px = ar 4 × ar 10

= a 2 r 4 + 10

Therefore, q 2 = ps

9. If a, a 2 + 2 and a 3  + 10 are in G.P., then find the values(s) of a.

(a 2 + 2) 2 = a(a 3 + 10)

a 4 + 4 = a 4 + 10a

4a 2 – 10a + 4 = 0

2a 2 – 5a + 2 = 0

2a 2 – a – 4a + 2 = 0

a(2a – 1) – 2(2a – 1) = 0

(2a – 1) (a – 2) = 0

Then, 2a – 1 = 0

a = ½

a – 2 = 0

Therefore, a = 2 or a= ½

10. Find the geometric progression whose 4 th term is 54 and the 7 th term is 1458.

4 th term a 4 = 54

7 th term a 7 = 1458

ar 6 = 1458

Now dividing we get,

ar 6 /ar 3 = (1458/54)

r 6 – 3 = 27

Then, ar 3 = 54

a × 27 = 54

Therefore G.P. is 2, 6, 18, 54, …

11. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

The sum of first three terms of a G.P. is 39/10

The product of first three terms of a G.P. is 1

Let us assume that a be the first term and ‘r’ be the common ratio,

And also assume that, three terms of the G.P. is a/r, a, ar,

The sum of three terms = (a/r) + a + ar = 39/10

Take out ‘a’ as common then, we get

a(1/r + 1 + r) = 39/10 … [equation (i)]

Now, product of three terms = (a/r) × a × ar = 1

a 3 r/r = 1

Substitute the value of ‘a’ in equation (i),

1(1/r + 1 + r) = 39/10

(1 + r + r 2 )/r = 39/10

By cross multiplication we get,

10(1 + r + r 2 )/r = 39r

10 + 10r + 10r 2 = 39r

Transposing 39r from right hand side to left hand side it becomes – 39r,

10 + 10r + 10r 2 – 39r = 0

10r 2 – 29r + 10 = 0

10r 2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(2r – 5) (5r – 2) = 0

So, 2r – 5 = 0

5r – 2 = 0

Therefore, r = 5/2 or 2/5

Then the terms if r = 5/2 are, 1, 5/2, 25/4, …

The terms if r = 2/5 are, 1, 2/5, 4/25, …

12. Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.

The sum of first three terms of a A.P. is 15

Let us assume three numbers are a – d, a, a + d.

The sum of three terms = a – d + a + a + d = 15

Then, adding 1, 4, 19 in the terms

The numbers become, a – d + 1, a + 4, a + d + 19

Therefore, b 2 = ac

(a + 4) 2 = (a – d + 1) (a + d + 19)

Simplify the above terms,

a 2 + 8a + 16 = a 2 + ad + 19a – ad – a 2 – 19d + a + d + 19

a 2 + 8a + 16 = a 2 – d 2 – 18d + 20a + 19

8a + 16 = 20a – 18d – d 2 + 19

8a + 16 – 20a + 18d + d 2 – 19 = 0

d 2 + 18d – 12a – 3 = 0

d 2 + 18d – (12 × 5) – 3 = 0

d 2 + 18d – 60 – 3= 0

d 2 + 18d – 63 = 0

d 2 + 21d – 3d – 63 = 0

d(d + 21) – 3(d + 21) = 0

(d + 21) (d – 3) = 0

So, d + 21 = 0

d = – 21

d – 3 = 0

Then the terms if d = 3 and a = 5,

Then G.P. 5 – 3 = 2, 5, 5 + 3 = 8

The terms if d = – 21 are 5 – (-21) = 5 + 21 = 26, 5, 5 – 21 = – 16

Exercise 9.5

1. Find the sum of:

(i) 20 terms of the series 2 + 6 + 18 + …

Common ratio r = 6/2 = 3

Number of terms n = 20

So, S 20 = a(r n – 1)/r – 1

= 2(3 20 – 1)/3 – 1

= 2(3 20 – 1)/2

= 3 20 – 1

Therefore, S 20 = 3 20 – 1

(ii) 10 terms of series 1 + √3 + 3 + …

Common ratio r = √3/1 = √3

Number of terms n = 10

So, S 10 = a(r n – 1)/r – 1

= 1((√3) 10 – 1)/ √3 – 1

Multiplying (√3 + 1) for both numerator and denominator we get,

= ((√3 10 – 1) (√3 + 1))/ ((√3 – 1) (√3 + 1)

= (3 5 – 1) (√3 + 1))/3 – 1 … [by rationalizing the denominator]

= ((243 – 1)( √3 + 1))/2

= 242(√3 + 1)/2

= 121(√3 + 1)

Therefore, S 10 = 121(√3 + 1

(iii) 6 terms of the GP 1, -2/3, 4/9, …

Common ratio r = -2/3 × 1= -2/3

Number of terms n = 6

So, S 6 = a(r n – 1)/r – 1

= 1[1 – (-2/3) 6 ]/(1 + (2/3))

= (3/5) (1 – (-2 6 /3 6 ))

= (3/5) (1 – (64/729))

= (3/5) ((729 – 64)/729)

= 3/5 × (665/729)

(iv) 5 terms and n terms of the series 1 + 2/3 + 4/9 + …

Common ratio r = 2/3 × 1= -/3

Number of terms n = 5

So, S n = a(1 – r n )/1 – r

= 1[1 – (2/3) n ]/(1 – 2/3)

S n = 3[1 – (2/3) n ]

Then, S 5 = 3[1 – (2/3) 5 ]

= 3[1 – (32/243)]

= 3((243 – 32)/243)

2. Find the sum of the series 81 – 27 + 9 … – 1/27

First term a = 81

Last term l = -1/27

S n = (a – lr)/(l – r)

= [81 + ((1/27) × (-1/3)]/[1 + (1/3)]

= [(81 – (1/81))]/(4/3)

= (6561 – 1)/[81 × (4/3)]

= (6560 × 3)/(81 × 4)

3. The n th term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.

The n th term of a G.P. T n = 128

The sum of its n terms S n = 255

Common ratio r = 2

We know that, T n = ar n – 1

128 = a2 n – 1

a = 128/2 n – 1 … [equation (i)]

Also we know that, S n = a(r n – 1)/(r – 1)

255 = a(2 n – 1)/(2 – 1)

255 = a(2 n – 1)

a = 255/(2 n  – 1) … [equation (ii)]

Now, consider both the equation(i) and equation (ii)

255/(2 n – 1) = 128/(2 n – 1 )

255 × 2 n – 1 = 128(2 n  – 1)

255 × 2 n – 1 = 128 × 2 n – 128

(255 × 2 n )/2 = 128 × 2 n – 128

255 × 2 n = 256 × 2 n – 256

256 × 2 n – 255 × 2 n = 256

By simplification,

2 n  = 256

2 n  = 2 8

By comparing both LHS and RHS, we get,

Then, 128 = a2 7

128 = a × 128

a = 128/128

Therefore, the value of a is 1.

4. (i) How many terms of the G.P. 3, 3 2 , 3 3 , … are needed to give the sum 120?

Terms of the G.P. 3, 3 2 , 3 3 , …

Sum of the terms = 120

We know that, S n = a(r n – 1)/r – 1 = 120

3(3 n – 1)/(3 – 1) = 120

3(3 n – 1)/2 = 120

3 n  – 1 = (120 ×2)/3

3 n – 1 = 240/3

3 n – 1 = 80

3 n = 80 +1

Therefore, n = 4

(ii) How many terms of the G.P. 1, 4, 16, … must be taken to have their sum equal to 341?

Terms of the G.P. 1, 4, 16, …

Sum of the terms = 341

We know that, S n = a(r n – 1)/r – 1 = 341

1(4 n – 1)/(4 – 1) = 341

1(4 n – 1)/3 = 341

4 n  – 1 = (341 × 3)

4 n – 1 = 1023

4 n = 1023 + 1

Therefore, n = 5

5. How many terms of the 2/9 – 1/3 + ½ + … will make the sum 55/72?

Terms of G.P. is 2/9 – 1/3 + ½ + …

Sum of the terms = 55/72

The first term a = 2/9

r = -1/3 ÷ 2/9 = (-1/3) × (9/2) = – 3/2

We know that, S n = a(r n – 1)/r – 1 = 55/72

1 – (-3/2) n = (55/72) × (5/2) × (9/2)

(1 – (-1)) n (3/2) n = 275/32

1 + 1(3/2) n = 275/32

(3/2) n = 275/32 – 1

(3/2) n = (275 – 32)/32

(3/2) n = 243/32

(3/2) n = (3/2) 5

6. The 2 nd and 5 th terms of a geometric series are -½ and sum 1/16, respectively. Find the sum of the series up to 8 terms.

a 2 = -½

We know that, a 2 = ar n – 1

= ar 2 – 1

a 2 = ar = -½ … [equation (i)]

a 5 = ar 4 = 1/16 … [equation (ii)]

Now, dividing equation (ii) by (i) we get,

r 3 = 1/16 ÷ (-½)

= (1/16) × (-2)

r 3 = (-1/2) 3

So, r = -½

ar = -½

a × (-½) = -½

a = – ½ × (-2/1)

Therefore, a = 1 and r = -½

Then, S 8 = a(1 – r n )/(1 – r)

= 1[1 – (-1/2) 8 ]/(1 + ½)

= [1 – (1/256)]/(3/2)

= (255/256) × (2/3)

= (510/768)

7. The first term of G.P. is 27 and 8 th term is 1/81. Find the sum of its first 10 terms.

First term a = 27

8 th term a 8 = 1/81

a 8 = ar 8 – 1 = 1/81

a 8 = ar 7 = 1/81

ar 7 = 1/81

27r 7 = 1/81

r 7 = 1/(81 × 27)

r 7 = 1/2187

r 7 = 1/(3 7 )

So, S 10 = a(1 – r n )/(1 – r)

= 27[1 – (1/3) 10 ]/(1 – 1/3)

= 27[1 – (1/3 10 )]/((3 – 1)/3)

= ((27 × 3)/2) [1 – 1/3 10 ]

= (81/2) [1 – 1/3 10 ]

8. Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.

Common ratio r = 3

Last term = 486

Sum of the terms = 728

We know that, S n = a(r n – 1)/(r – 1)

= a(3 n – 1)/(3 – 1) = 728

a(3 n – 1)/2 = 728

a(3 n – 1) = 728 × 2

a(3 n – 1) = 1456 … [equation (i)]

Then, last term = ar n – 1

486 = a × 3 n – 1

486 = a(3 n /3)

486 × 3 = a3 n

1458 = a3 n … [equation (ii)]

Consider equation (i), a(3 n – 1) = 1456

a3 n – a = 1456

Substitute the value of a3 n in equation (i),

1458 – a = 1456

a = 1458 – 1456

Therefore, the first term a is 2.

9. In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio.

First term a is = 7

Then, last term is = 448

We know that, last term = ar n – 1

7r n -1 = 448

r n – 1 = 448/7

r n – 1 = 64 … [equation (i)]

So, sum = a(r n – 1)/(r – 1) = 889

7(r n – 1)/(r – 1) = 889

(r n – 1)/(r – 1) = 889/7

(r n – 1)/(r – 1) = 127 … [equation (ii)]

Consider the equation (i),

r n /r = 64

Now substitute the value of r n in equation (ii),

(64r – 1)/(r – 1) = 127

64r – 1 = 127r – 127

127r – 64r = -1 + 127

Therefore, common ratio = 2

10. Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.

S 7 = a(3 n – 1)/(3 – 1)

2186 = a(3 n  – 1)/2

(2186 × 2) = a(3 7 – 1)

(4372) = a(2187 – 1)

4372 = a2186

a = 4372/2186

Then, a 3 = ar 3 – 1

= 2 × 3 2

= 2 × 9

11. If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio.

First term of a G.P. is a = 5

The sum of first three terms is S 3 = 31/5

S 3 = a(r 3 – 1)/(r – 1)

31/5 = 5(r 3 – 1)/(r – 1)

31/(5 × 5) = (r 3 – 1)/(r – 1)

31/25 = (r 3 – 1)/(r – 1)

(r – 1) (r 2 + r + 1)/(r – 1) = 31/25

r 2 + r + 1 = 31/25

25(r 2 + r + 1) = 31

25r 2 + 25r + 25 = 31

Transposing 31 from right hand side to left hand side it becomes – 31,

25r 2 + 25r + 25 – 31 =

25r 2 + 25r – 6 = 0

25r 2 + 30r – 5r – 6 = 0

5r(5r + 6) – 1(5r + 6) = 0

(5r – 1) (5r + 6) = 0

Take 5r – 1 = 0

or 5r + 6 = 0

Therefore, common ratio r = 1/5 or -6/5.

Chapter – test

1. Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.

First term a = – 5

Then the first four terms are = – 5 + (-3) = -5 – 3 = – 8

-8 + (-3) = – 8 – 3 = -11

– 11 + (-3) = – 11 – 3 = – 14

Therefore, first four terms are -5, -8, -11 and -14.

2. Verify that each of the following lists of numbers is an A.P., and the write its next three terms:

(i) 0, ¼, ½, ¾, …

First term a = 0

Common difference = ¼ – 0 = ¼

So, next three numbers are ¾ + ¼ = 4/4 = 1

1 + ¼ = (4 + 1)/4 = 5/4

5/4 + ¼ = 6/4 = 3/2

Therefore, the next three term are 1, 5/4 and 3/2.

(ii) 5, 14/3, 13/3, 4, …

Common difference = 14/3 – 5 = (14 – 15)/3 = -1/3

So, next three numbers are 4 + (-1/3) = (12 – 1)/3 = 11/3

11/3 + (-1/3) = (11 – 1)/3 = 10/3

10/3 + (-1/3) = (10 – 1)/3 = 9/3 = 3

Therefore, the next three term are 11/3, 10/3 and 3.

3. The n th term of an A.P. is 6n + 2. Find the common difference.

n th term is 6n + 2

So, T n = 6n + 2

T 1 = (6 × 1) + 2 = 6 + 2 = 8

T 2 = (6 × 2) + 2 = 12 + 2 = 14

T 3 = (6 × 3) + 2 = 18 + 2 = 20

T 4 = (6 × 4) + 2 = 24 + 2 = 26

Therefore, A.P. is 8, 14, 20, 26, …

So, common difference d = 14 – 8 = 6

4. Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16 th term and the n th .

The first term a = 9

Then, difference d = 12 – 9 = 3

15 – 12 = 3

18 – 15 = 3

Therefore, common difference d = 3

= 9 + (n – 1)3

= 9 + 3n – 3

So, T 16 = a + (n – 1)d

= 9 + (16 – 1)3

= 9 + (15)(3)

5. Find the 6 th term from the end of the A.P. 17, 14, 11, …, – 40.

Common difference = 14 – 17 = – 3

Last term l = – 40

L = a + (n – 1)d

-40 = 17 + (n – 1)(-3)

-40 – 17 = -3n + 3

– 57 – 3 = -3n

Therefore, 6 th term form the end = l – (n – 1)d

= – 40 – (6 – 1)(-3)

= – 40 – (5)(-3)

= – 40 + 15

= – 25

6. If the 8 th term of an A.P. is 31 and the 15 th term is 16 more than its 11 th term, then find the A.P.

a 15 = the 15 th term is 16 more than its 11 th term = a 11 + 16

we know that, a n = a + (n – 1)d

So, a 8 = a + 7d = 31 … [equation (i)]

a 15 = a + 14d = a + 10d + 16

14d – 10d = 16

Now substitute the value of d in equation (i) we get,

a + (7 × 4) = 31

a + 28 = 31

a = 31 – 28

So, 3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15

Therefore, A.P. is 3, 7, 11, 15, …

7. The 17 th term of an A.P. is 5 more than twice its 8 th term. If the 11 th term of the A.P. is 43, then find the w th term.

a 17 = 5 more than twice its 8 th term = 2a 8 + 5

We know that, a 11 = a + 10d = 43 … [equation (i)]

a 17 = 2a 8 + 5

a + 16d = 2(a + 7d) + 5

a + 16d = 2a + 14d + 5

2a – a = 16d – 14d – 5

a = 2d – 5 … [equation (ii)]

Now substitute the value of a in equation (i) we get,

2d – 5 + 10d = 43

12d = 43 + 5

To find out the value of a substitute the value of d in equation (i)

a + (10 × 4) = 43

a + 40 = 43

a = 43 – 40

Then, a n = a + (n – 1)d

= 3 + 4(n – 1)

= 3 + 4n – 4

= 4n – 1

8. The 19 th term of an A.P. is equal to three times its 6 th term. If its 9 th term is 19, find the A.P.

a 19 = 19 th term of an A.P. is equal to three times its 6 th term = 3a 6

As we know, a n = a + (n – 1)d

a 9 = a + 8d = 19 … [equation (i)]

Then, a 19 = 3(a + 5d)

a + 18d = 3a + 15d

3a – a = 18d – 15d

(3/2)d + 8d = 19

(3d + 16d)/2 = 19

(19/2)d = 19

d = (19 × 2)/19

a + 8d = 19

a + (8 × 2) = 19

a + 16 = 19

a = 19 – 16

Therefore, A.P. is 3, 5, 7, 9, …

9. If the 3 rd and the 9 th terms of an A.P. are 4 and – 8, respectively, then which term of this A.P. is zero?

a 9 = – 8

We know that, a 3 = a + 2d = 4 … [equation (i)]

a 9 = a + 8d = -8 … [equation (ii)]

Now, subtracting equation (i) from equation (ii)

(a + 8d) – (a + 2d) = -8 – 4

a + 8d – a – 2d = -12

a + (2 × (-2)) = 4

a – 4 = 4

let us assume n th term be zero, then

a + (n – 1)d = 0

8 + (n – 1)(-2) = 0

-2n + 2 = -8

-2n = -8 – 2

Therefore, 0 will be the fifth term.

10. Which term of the list of numbers 5, 2, – 1, – 4, … is – 55?

n th term = -55

Common difference d = 2 – 5 = – 3

– 55 = 5 + (n – 1)(-3)

-55 – 5 = – 3n + 3

-60 – 3 = -3n

Therefore, -55 is the 21 st term.

11. The 24 th term of an A.P. is twice its 10 th term. Show that its 72 nd term is four times its 15 th term.

The 24 th term of an A.P. is twice its 10 th term = a 24 = 2a 10

We have to show that, 72 nd term is four times its 15 th term = a 72 = 4a 15

We know that, a 24 = a + 23d = 2a 10

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

2a – a = 23d – 18d

a = 5d … [equation (i)]

a 72 = 4a 15

a + 71d = 4(a + 14d)

Substitute the value of a we get,

5d + 71d = 4(5d + 14d)

76d = 4(19d)

Therefore, it is proved that 72 nd term is four times its 15 th term.

12. Which term of the list of numbers 20, 19¼, 18½, 17¾, … is the first negative term?

First term a = 20

Common difference d = 19¼ – 20 = 77/4 – 20 = (77 – 80)/4 = -¾

a n = 20 + (n – 1) (-¾)

a n = 20 – ¾n + ¾

a n = 20 + ¾ – ¾n

a n = (80 + 3)/4 – ¾n

a n = 83/4 – ¾n < 0

83/4 < ¾n

83/3 < n

Therefore, 28 th is the first negative term.

13. How many three digit numbers are divisible by 9?

The three digits numbers which are divisible by 9 are 108, 117, 126, …, 999

Then, first term a = 108

Common difference = 9

Last term = 999

We know that, l = a n = a + (n – 1)d

999 = 108 + (n – 1)9

999 – 108 = 9n – 9

891 + 9 = 9n

Therefore, there are 100 three digits numbers.

14. The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.

The sum of three numbers in A.P. = – 3

The product of three numbers in A.P. = 8

Now adding 3 numbers = a – d + a + a + d = – 3

So, (a – d) × (a) × (a + d) = 8

a(a 2 – d 2 ) = 8

-1 ((-1) 2 – d 2 = 8

1 – d 2 = 8/-1

1 – d 2 = -8

d 2 = 8 + 1

d = √9

d = ±3

Therefore, the numbers are if d = 3 (a – d) = – 1 – 3 = – 4

(a + d) = – 1 + 3 = 2

The numbers are (a – d) = – 1 – (-3) = – 1 + 3 = 2

(a + d) = -1 + (-3) = -1 – 3 = -4

Therefore, the numbers -4, -1, 2,… and 2, -1, -4,… are in A.P.

15. The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.

The angles of a quadrilateral are in A.P.

Greatest angle is double of the smallest angle

Let us assume the greatest angle of the quadrilateral is a + 3d,

Then, the other angles are a + d, a – d, a – 3d

So, a – 3d is the smallest

Therefore, a + 3d = 2(a – 3d)

a + 3d = 2a – 6d

6d + 3d = 2a – a

9d = a … [equation (i)]

We know that the sum of all angles of quadrilateral is 360 o .

a – 3d + a – d + a + d + a+ 3d = 360 o

Now, substitute the value of a in equation (i) we get,

Substitute the value of a and d in assumed angles,

Greatest angle = a + 3d = 90 + (3 × 10) = 90 + 30 = 120 o

Then, other angles are = a + d = 90 o + 10 o = 100 o

a – d = 90 o – 10 o  = 80 o

a – 3d = 90 o  – (3 × 10) = 90 – 30 = 60 o

Therefore, the angles of quadrilateral are 120 o , 100 o , 80 o and 60 o .

16. Find the sum of first 20 terms of an A.P. whose n th term is 15 – 4n.

n th term is 15 – 4n

So, a n = 15 – 4n

a 1 = 15 – (4 × 1) = 15 – 4 = 11

a 2 = 15 – (4 × 2) = 15 – 8 = 7

a 3 = 15 – (4 × 3) = 15 – 12 = 3

a 4 = 15 – (4 × 4) = 15 – 16 = – 1

Then, a 20 = 15 – (4 × 20) = 15 – 80 = -65

So, 11, 7, 3, -1, … -65 are in A.P.

Therefore, first term a = 11

Common difference = -4

S 20 = (n/2) [2a + (n – 1)d]

= (20/2) [(2 × 11) + (20 – 1)(-4)]

= 10 [22 – (19) (-4)]

= 10 [22 – 76]

Therefore, the sum of first 20 terms of an A.P. is -540.

17. Find the sum : 18 + 15½ + 13 + … + (-49½)

First term a = 18

Common difference d = 15½ – 18

= 31/2 – 18

= (31 – 36)/2

Last term = -49½ = -99/2

-99/2 = 18 + (n – 1)(-5/2)

(-99/2) – (18/1) = (n – 1)(-5/2)

(-99 – 36)/2 = (-5/2)(n – 1)

(-135/2) = (-5/2) (n – 1)

(-135/2) × (-2/5) = n – 1

-135/-5 = n – 1

27 = n – 1

Then, S n = (n/2) [2a + (n – 1)d]

S 28 = (28/2) [(2 × 18) + (28 – 1)(-5/2)]

S 28 = 14[36 + (27 × (-5/2))]

S 28 = 14[36 – (135/2)]

S 28 = 14 [(72 – 135)/2]

S 28 = 14 (-63/2)

S 28 = – 441

18. (i) How many terms of the A.P. – 6, (-11/2), – 5, … make the sum – 25?

Terms of the A.P. is -6, (-11/2) – 5, …

The first term a = -6

Common difference d = (-11/2) – (-6)

= (-11/2) + 6

= (-11 + 12)/2

The terms are make the sum – 25

Then, S n = (n/2)(2a + (n – 1)d)

-25 = (n/2) [(2 × (-6)) + (n – 1) (½)]

(-25 × 2) = n [-12 + ½n – ½]

-50 = n [(-25/2) + (½n)]

½n 2 – (25/2)n + 50 = 0

n 2 – 25n + 100 = 0

n 2 – 5n – 20n + 100 = 0

n(n – 5) – 20(n – 5) = 0

(n – 5) (n – 20) = 0

So, n – 5 = 0

or n – 20 = 0

Therefore, number of terms are 5 or 20.

(ii) Solve the equation 2 + 5 + 8 + … + x = 155

Last term = x

Common difference d = 5 – 2 = 3

Then, sum of the terms = 155

x = 2 + (n – 1)3

x = 2 + 3n – 3

x = 3n – 1 … [equation (i)]

We know that, S n = (n/2) [2a + (n – 1)d]

155 = (n/2) [(2 × 2) + (n – 1) × 3]

155 × 2 = n[4 + 3n – 3]

310 = n(3n + 1)

310 = 3n 2 + n

3n 2 + n – 310 = 0

3n 2 – 30n + 31n – 310 = 0

3n(n – 10) + 31(n – 10) = 0

(n – 10) (3n + 31) = 0

So, n – 10 = 0

or 3n + 31 = 0

negative is not possible.

Therefore, n = 10

x = 3n – 1

= (3 × 10) – 1

= 30 – 1

19. If the third term of an A.P. is 5 and the ratio of its 6 th term to the 10 th term is 7 : 13, then find the sum of first 20 terms of this A.P.

The third term of an A.P. a 3 = 5

The ratio of its 6 th term to the 10 th term a 6 : a 10 = 7 : 13

a 3 = a + (3 – 1)d = 5

= a + 2d = 5 … [equation (i)]

Then, a 6 /a 10 = 7/13

(a + 5d)/(a + 9d) = 7/13

13(a + 5d) = 7(a + 9d)

13a + 75d = 7a + 63d

13a – 7a + 65d – 63d = 0

6a + 2d = 0

Divide by 2 on both side we get,

d = -3a … [equation (ii)]

Substitute the value of d in equation (i),

a + 2(-3a) = 5

a – 6a = 5

Now substitute the value of a in equation (ii),

Then, sum of first 20 terms,

= (n/2) [2a + (n – 1)d]

= (20/2)[(2 × (-3)) + (2- – 1)3]

= 10[-2 + 57]

20. The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25 th term.

The sum of first 14 terms of an A.P. = 1505

25 th term = ?

S 14 = (n/2) [2a + (n – 1)d]

1505 = (14/2) [(2 × 10) + (14 – 1)d]

1505 = 7[20 + 13d]

1505/7 = 20 + 13d

215 = 20 + 13d

13d = 215 – 20

a 25 = 10 + (25 – 1)(15)

= 10 + (24)15

21. Find the geometric progression whose 4 th term is 54 and 7 th term is 1458.

The geometric progression whose 4 th term a 4 = 54

The geometric progression whose 7 th term a 7 = 1458

We know that, a n = ar n – 1

a 4 = ar 3 = 54

a 7 = ar 6 = 1458

By dividing both we get,

ar 6 /ar 3 = 1458/54

To find out a, consider ar 3 = 54

a(3) 3 = 54

Therefore, a = 2, r = 3

So, G.P. is 2, 6, 18, 54,…

22. The fourth term of a G.P. is the square of its second term and the first term is – 3. Find its 7 th term.

The fourth term of a G.P. is the square of its second term = a 4 = (a 2 ) 2

The first term a 1 = – 3

Now, ar 3 = (ar) 2

ar 3 = a 2 r 2

r 3 /r 2 = a 2 /a

r 3 – 2 = a 2 – 1 … [from a m /a n  = a m – n ]

a 7 = ar 7 – 1

= -3 × (-3) 6

= -3 × 729

Therefore, the 7 th term a 7 = -2187

23. If the 4 th , 10 th and 16 th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.

Now, we have to show that x, y and z are in G.P.

a n = ar n – 1

a 4 = ar 3 = x

a 10 = ar 9 = y

a 16 = ar 15 = z

x, y, z are in G.P.

If y 2 = xy

Substitute the value of x and y,

y 2 = (ar 9 ) 2

y 2 = a 2 r 18

Then, xz = ar 3 × ar 15

= a 1 + 1 r 3 + 15 … [from a m × a n = a m + n ]

So, y 2 = xy

Therefore, it is proved that x, y, z are in G.P.

24. How many terms of the G.P. 3, 3/2, ¾ are needed to give the sum 3069/512?

Sum of the terms S n = 3069/512

Common ratio r = (3/2)/3

= (3/2) × (1/3)

We know that, S n = a(1 – r n )/(1 – r)

(3069/512) = 3[1 – (½) n ]/ (1 – ½)

(3069/512) = (2 × 3) [1 – (½) n ]

1 – (½) n = 3069/(512 × 6)

1 – (½) n = 1023/1024

(½) n = 1 – (1023/1024)

(½) n = (1024 – 1023)/1024

(½) n = 1/1024

(½) n = (½) 10

By comparing both LHS and RHS,

Therefore, there are 10 terms are in the G.P.

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• Polynomials Class 9 Assertion Reason Questions Maths Chapter 2

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Assertion Reason questions are the new question format that is introduced in CBSE board. The resources for assertion reason questions are very less. So, to help students we have created chapterwise assertion reason questions for class 9 maths. In this article, you will find assertion reason questions for CBSE Class 9 Maths Chapter 2 Polynomials. It is a part of Assertion Reason Questions for CBSE Class 9 Maths Series.

Table of Contents

Assertion Reason Questions on Polynomials

Questions :

Q. 1. Assertion (A): If $p(x)=x^2-4 x+3$, then 3 and 1 are the zeroes of the polynomial $p(x)$. Reason (R): Number of zeroes of a polynomial cannot exceed its degree.

Q. 2. Assertion (A): The degree of the polynomial $(x-2)(x-3)(x+4)$ is 4. Reason (R): The number of zeroes of a polynomial is the degree of that polynomial.

Q. 3. Assertion (A): Factorisation of the polynomial $\sqrt{3} x^2+11 x+6 \sqrt{3}$ is $(\sqrt{3} x+2)(x+\sqrt{3})$. Reason (R): Factorisation of the polynomial $35 y^2+13 y-12$ is $(7 y-3)(5 y+4)$.

1. (b) Assertion (A): Given, $p(x)=x^2-4 x+3$

$$ \begin{aligned} \Rightarrow \quad p(x) & =x^2-(3+1) x+3 \\ & =x^2-3 x-x+3 \\ & =x(x-3)-1(x-3) \\ & =(x-1)(x-3) \end{aligned} $$

For finding the zeroes, put $p(x)=0$

$$ \therefore(x-1)(x-3)=0 \Rightarrow x=1,3 $$

So, Assertion (A) is true. Reason (R): It is true to say that the number of zeroes of a polynomial cannot exceed its degree. Hence, both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

2. (d) Assertion (A): $p(x)=(x-2)(x-3)(x+4)$

$$ \begin{aligned} & =(x-2)\left[x^2+4 x-3 x-12\right] \\ & =(x-2)\left(x^2+x-12\right) \\ & =x^3+x^2-12 x-2 x^2-2 x+24 \\ p(x) & =x^3-x^2-14 x+24 \end{aligned} $$

So, degree of $p(x)=3$. Hence, Assertion (A) is false, but Reason (R) is true.

3. (d) Assertion (A):

$$ \begin{aligned} \left(\sqrt{3} x^2+11 x+6 \sqrt{3}\right) & =\sqrt{3} x^2+9 x+2 x+6 \sqrt{3} \\ & =\sqrt{3} x(x+3 \sqrt{3})+2(x+3 \sqrt{3}) \\ & =(x+3 \sqrt{3})(\sqrt{3} x+2) \end{aligned} $$

So, Assertion (A) is false. Reason (R):

$$ \begin{aligned} \left(35 y^2+13 y-12\right) & =35 y^2+28 y-15 y-12 \\ & =7 y(5 y+4)-3(5 y+4) \\ & =(5 y+4)(7 y-3) \end{aligned} $$

So, Reason (R) is true.

Number Systems Class 9 Assertion Reason Questions Maths Chapter 1

Topics from which assertion reason questions may be asked.

• Definition of a polynomial in one variable, with examples and counter examples.
• Coefficients of a polynomial
• Terms of a polynomial and zero polynomial.
• Degree of a polynomial.
• Constant, linear, quadratic and cubic polynomials.
• Monomials, binomials, trinomials.
• Factors and multiples.
• Zeros of a polynomial.
• Remainder Theorem with examples.

Linear polynomial can be monomial or binomial. Quadratic polynomial can be monomial, binomial or trinomial. The degree of a zero polynomial is not defined.

Assertion reason questions from the above given topic may be asked.

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Frequently Asked Questions (FAQs) on Polynomials Assertion Reason Questions Class 9

Q1: what are assertion reason questions.

A1: Assertion-reason questions consist of two statements: an assertion (A) and a reason (R). The task is to determine the correctness of both statements and the relationship between them. The options usually include: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true. or A is false, and R is also false.

Q2: Why are assertion reason questions important in Maths?

A2: Students need to evaluate the logical relationship between the assertion and the reason. This practice strengthens their logical reasoning skills, which are essential in mathematics and other areas of study.

Q3: How can practicing assertion reason questions help students?

A3: Practicing assertion-reason questions can help students in several ways: Improved Conceptual Understanding:  It helps students to better understand the concepts by linking assertions with their reasons. Enhanced Analytical Skills:  It enhances analytical skills as students need to critically analyze the statements and their relationships. Better Exam Preparation:  These questions are asked in exams and practicing them can improve your performance.

Q4: What strategies should students use to answer assertion reason questions effectively?

A4: Students can use the following strategies: Understand Each Statement Separately:  Determine if each statement is true or false independently. Analyze the Relationship:  If both statements are true, check if the reason correctly explains the assertion.

Q5: What are common mistakes to avoid when answering Assertion Reason questions?

A5: Common mistakes include: Not reading the statements carefully and missing key details. Assuming the Reason explains the Assertion without checking the logical connection. Confusing the order or relationship between the statements. Overthinking and adding information not provided in the question.

Q6: H ow many types of polynomials are there?

A6: Polynomials are classified based on the number of terms they have: Monomial:  A polynomial with just one term. Binomial:  A polynomial with two terms. Trinomial:  A polynomial with three terms Multinomial:  A polynomial with more than three terms

Q7: What is the degree of a polynomial?

A7: The degree of a polynomial is the highest power of the variable in the polynomial.

Q8: What are the key concepts covered in Chapter 2 of CBSE Class 9 Maths regarding polynomials?

A8: Chapter 2 of CBSE Class 9 Maths covers concepts such as understanding polynomials and its types. (i) Types of polynomials (ii) Terms and coefficient of polynomials (iii) Zeroes of a polynomial (iv) Division algorithm (v) Remainder theorem (vi) Factor theorem (vii) Factorisation of quadratic polynomial

Q9: What are the important keywords for CBSE Class 9 Maths Polynomials?

A9: List of important keywords given below – Algebraic Expression:  Any expression that contains constants and variables, connected by some or all of the operations +, -, x, ÷. Polynomials:  An algebraic expression in which the variables involved have only non-negative integral powers. Polynomials in one Variable:  An algebraic expression which consist of only one type of variables in the entire expression. Degree of Polynomial:  Highest power of a variable in the polynomial. Constant Polynomial:  Polynomial of zero degree. Zero Polynomials:  A polynomial consisting of one term, namely zero. Zeroes of a Polynomial:  Let p(x) be a polynomial in one variable and ‘a’ be a real number such that the value of polynomial at x=a is zero i.e., p(a) = 0, then ‘a’ is said to be a zero polynomial of p(x). Remainder Theorem:  Let p(x) be a polynomial having degree 1 or more than 1 and let ‘a’ be any real number. If p(x) is divided by (x-a), then remainder is p(a).

Q10: Are there any online resources or tools available for practicing linear equations in one variable assertion reason questions?

A10: A9: We provide assertion reason questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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