# NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.2

Students can prepare for the exams by downloading the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. However, students should go through the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 available on the Extramarks’ website and mobile app.

The NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 provide the students with a clear understanding of all the important concepts and formulas. Furthermore, students can easily solve the Mathematics exercises and evaluate their answers with the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. Furthermore, with the support of the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2, Class 11 students can understand the proper way of answering Mathematics questions.

Chapter 9 of Class 11 is quite difficult, and students get confused about how to solve the practise questions. The NCERT books have various exercises for Chapter 9. Students can take help from the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 to solve the exercises.Extramarks offers live sessions on their website and mobile app where students can ask questions about the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2.Exercise 9.2

Class 11 Maths NCERT Solutions Chapter 9 Sequences and Series excellently help students with the exercise questions in the chapter. These solutions are precisely solved by Extramarks’ Experts who have taken a simple approach to make the concepts more smooth.

Arithmetic Progression, Geometric Progression, the Correlation between them, and other interesting topics are covered in Class 11 Maths Chapter 9. Students should prioritise practising NCERT solutions to gain a good grasp on these topics.

Students who are facing challenges in Chapter 9 can take guidance from the  NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. On the Extramarks’ website and mobile application, students can easily get the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2.

Students can download the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 offline in PDF format.

Often, students are unable to solve this exercise, so they can check the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 on the Extramarks’ website. The second exercise in Chapter 9 focuses on Arithmetic Progression and Arithmetic Mean. Arithmetic progression (A.P.) or arithmetic sequence is a number sequence in which the variation between consecutive terms is constant. The practise also includes Arithmetic Mean, which is described as the average of a set of numerical values determined by adding and dividing them by the number of terms in the set. NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 will help students learn and understand more about this exercise.

Sometimes students get tired while sitting on a laptop taking live lectures, students can easily download the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 in PDF format for offline access. Downloading the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 makes practise easy for students, and they can do it without sitting on a laptop for a longer period.

## NCERT Solutions for Class 11 Maths Chapters

Apart from concept revision, it is referred to consistently practice the exercise questions to score well in the Class 11 Mathematics exam. To effectively revise the chapters, students must study from the Mathematics revision notes. Extramarks’ website and mobile application help students download revision notes in PDF format.

#### Chapter 1: Sets

Sets are an important chapter in Mathematics that is used to categorise the concepts of relations and capacities. The first chapter of the Class 11 Maths NCERT book will go over sets, focusing on how to deal with the set hypothesis, finite and infinite sets, and so on.

#### Chapter 2: Relations and Functions

Students will learn about the special relationships between objects, functions, and the mathematical correspondence between two quantities in this chapter.

#### Chapter 3: Trigonometric Functions

Chapter 3 will guide the students through more complex trigonometric concepts, such as trigonometric ratios and functions. Students will also explore their characteristics.

#### Chapter 4: Principle of Mathematical Induction

The principle of Mathematical Induction is an exciting chapter. In this section, students will learn about Maths Induction and the Principle of Mathematical Induction. Students will then be required to apply those ethics to a variety of problems.

#### Chapter 5: Complex Numbers and Quadratic Equations

Chapter 5 will assist students in understanding complex numbers and problem solving based on them. It will also discuss the square roots of negative numbers, the Argand Plane, and quadratic equations.

#### Chapter 6: Linear Inequalities

In this section, students will discover inequalities, algebraic solutions to linear inequalities in one variable, graphical solutions to linear inequalities in two variables, and a wide range of other applications of linear inequalities.Chapter 7: Permutations and Combinations

This chapter will discuss permutations, combinations, permutations with distinct or non-distinct objects, Factorial notation, and other related topics.

#### Chapter 8: Binomial Theorem

Students will learn about the binomial theorem for positive integral indices, Pascal’s triangle, and how to apply the binomial theorem in a wide range of situations in Chapter 8.

#### Chapter 9: Sequences and Series

Students will learn about Arithmetic Progression, Geometric Progression, the General Term of a G.P, sum to n terms of a G.P, and other important topics in the chapter ‘Sequences and Series.’ It is an essential part of Class 11 Math. The exercise in Chapter 9 is very interesting. Moreover, students who are facing a few challenges in solving the 9.2 Class 11 can access the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2.

#### Chapter 10: Straight Lines

This chapter will introduce students to many important concepts related to straight lines. It will be divided into topics like the Slope of a line, the angle between two lines, Collinearity between two points, and many more such topics.

#### Chapter 11: Conic Sections

Chapter 11 will expand on these concepts and introduce students to conic segments such as parabolas and hyperbolas.

#### Chapter 12: Introduction to Three Dimensional Geometry

Students will learn about coordinate geometry in three dimensions in this chapter.

#### Chapter 13: Limits and Derivatives

Students will be introduced to the concept of derivatives and limits, along with the limits and derivatives of polynomials and trigonometric functions.

#### Chapter 14: Mathematical Reasoning

Statements, Negation of a Statement, Compound Statements, Quantifiers, Implications, Validating Statements, and many other concepts will be covered in Chapter 14.

Chapter 15: Statistics

Students will learn how to deal with data for specific purposes.

Chapter 16: Probability

Probability deals with the likelihood of the occurrence of an event.

### NCERT Solution Class 11 Maths of Chapter 9 Exercises

The NCERT solutions for all Class 11 subjects are available on Extramarks’ website and mobile application. In addition, the NCERT solutions can assist students in effectively preparing for the Class 11 final exams.Practising a large number of questions leads to finishing the exam on time. Students of all classes can access the NCERT Solutions for each subject via the Extramarks’ website and mobile application.

There are many exercises in Chapter 9, but the second exercise is quite difficult. Many students got confused while solving exercise 9.2 Class 11th. Students can get help from the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. Students can access the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 on the website and the mobile application of Extramarks.

The weightage of Class 11 chapter 9 is 8 marks, and the second exercise is one of the main exercises. Therefore, students should start preparing for the exam with the help of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2.

### All Important Topics and Formulae to Solve Exercise 9.2 — Class 11 NCERT Maths

There are many formulas in Chapter 9 of Class 11. Students should always do the exercises on a daily practise basis. Extramarks help students in solving exercise 9.2. The NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 allows students to understand the exercise.

Class 11 Maths has a lot of formulas; which can be challenging for the students. Moreover, solving exercise questions will help students  keep the formulas in mind. The exercises in Chapter 9 include questions about Arithmetic Mean, the average of a set of numerical values calculated by adding them all together and dividing by the number of terms in the set.

The NCERT textbook contains a plethora of questions for students to solve and practice. For example, practising with the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 will help students achieve high grades in their Class 11 exams. Students who are having difficulty with Chapter 9 can refer to the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. Students can easily get the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 from the Extramarks’ website and mobile application.

### NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.2

The Extramarks’ website and mobile application contain the necessary resources for achieving higher grades on board exams. Furthermore, while studying for the Mathematics exams, they should keep referring to the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2. Practising the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 can assist students in comprehending the topics covered in the chapter. Highly experienced teachers prepare these solutions following the most recent CBSE Syllabus.

Chapter 9 of Class 11 is quite difficult, and some students are unsure how to answer the question papers. A decent amount of questions in Chapter 9 can be found in the NCERT books. In addition, students can use the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 to solve the exercise questions. Extramarks offers live sessions on their website and mobile app where students can ask questions about the NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2.

Q.1 Find the sum of odd integers from 1 to 2001.

Ans.

$\begin{array}{l}\mathrm{The}\text{odd integers from 1 to 2001 are:}\\ 1,\text{\hspace{0.17em}}3,5,7,9,11,13,...,1999,2001\\ \mathrm{a}=1,\text{\hspace{0.17em}}\mathrm{d}=3-1=2\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}}2001=1+\left(\mathrm{n}-1\right)2\\ 2001-1=\left(\mathrm{n}-1\right)2\\ \mathrm{n}-1=\frac{2000}{2}\\ \text{}\mathrm{n}=1000+1\\ =1001\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1001}=\frac{1001}{2}\left(1+2001\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1001}=\frac{1001}{2}\left(2002\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1001\left(1001\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1002001\\ \mathrm{Thus},\text{​​ the sum of odd integers from 1 to 2001 is 1002001.}\end{array}$

Q.2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Ans.

$\begin{array}{l}\mathrm{The}\text{multiples of 5 from 100 to 1000 are:}\\ 105,\text{\hspace{0.17em}}105,110,\text{\hspace{0.17em}}115,\text{\hspace{0.17em}}120,\text{\hspace{0.17em}\hspace{0.17em}}125,...,995\\ \mathrm{a}=100,\text{\hspace{0.17em}}\mathrm{d}=105-100=5\\ \mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ 995=105+\left(\mathrm{n}-1\right)5\\ 995-105=\left(\mathrm{n}-1\right)5\\ \mathrm{n}-1=\frac{890}{5}\\ \text{}\mathrm{n}=178+1\\ \text{}=179\\ \text{}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{}{\mathrm{S}}_{179}=\frac{179}{2}\left(105+995\right)\\ =\frac{179}{2}×\left(1100\right)\\ =98450\\ \mathrm{Thus},\text{the sum of the natural numbers which are multiple of 5}\\ \text{between 100 to 1000 is}98450.\end{array}$

Q.3 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{First}\text{term of A.P.}=\text{2}\\ \text{Let common difference}=\mathrm{d}\\ \mathrm{According}\text{to given condition:}\\ \text{Sum of first 5 terms}=\frac{1}{4}\left(\mathrm{Sum}\text{of next 5 terms}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{5}=\frac{1}{4}\left({\mathrm{S}}_{10}-{\mathrm{S}}_{5}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{S}}_{5}={\mathrm{S}}_{10}-{\mathrm{S}}_{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5{\mathrm{S}}_{5}={\mathrm{S}}_{10}\end{array}$

$\begin{array}{l}⇒5\left[\frac{5}{2}\left\{2×2+\left(5-1\right)\mathrm{d}\right\}\right]=\left[\frac{10}{2}\left\{2×2+\left(10-1\right)\mathrm{d}\right\}\right]\\ \left[\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{25}{2}\left(4+4\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25\left(2+2\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50+50\mathrm{d}=20+45\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{d}=20-50\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=-\frac{30}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-6\\ \mathrm{Now},\text{sum of 20 terms.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{20}}=2+\left(20-1\right)×-6\left[\because {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2-114\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-112\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term is}-\text{112.}\end{array}$

Q.4

$\begin{array}{l}\mathbf{\text{How many terms of the A.P.\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\mathbf{\text{6,}}\mathbf{-}\frac{\mathbf{\text{11}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{-}\mathbf{\text{5,}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em} are needed to give}}\\ \mathbf{\text{the sum}}\mathbf{-}\mathbf{\text{25?}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{The}\text{given A.P. is}-6,-\frac{11}{2},-5,...\text{\hspace{0.17em}}\\ \mathrm{Let}\text{the sum of n terms of A.P. is}-\text{25.Then,}\end{array}$

$\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\\ \mathrm{a}=-6\\ \mathrm{d}=-\frac{11}{2}-\left(-6\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{11}{2}+6=\frac{1}{2}\\ -25=\frac{\mathrm{n}}{2}\left\{2×-6+\left(\mathrm{n}-1\right)×\frac{1}{2}\right\}\left[\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\\ -25=\frac{\mathrm{n}}{2}\left\{-12+\left(\mathrm{n}-1\right)×\frac{1}{2}\right\}\\ -100=\mathrm{n}\left(-24+\mathrm{n}-1\right)\\ -100=-25\text{\hspace{0.17em}}\mathrm{n}+{\mathrm{n}}^{2}\\ {\mathrm{n}}^{2}-25\text{\hspace{0.17em}}\mathrm{n}+100=0\\ \left(\mathrm{n}-20\right)\left(\mathrm{n}-5\right)=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=20,\text{\hspace{0.17em}\hspace{0.17em}}5\\ \mathrm{Thus},\text{the number of terms of A.P. is 5 or 20.}\end{array}$

Q.5

$\begin{array}{l}{\mathbf{\text{In​​ an A.P., if p}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{p}}}{\mathbf{\text{​​ and q}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{q}}}\mathbf{\text{, prove that the sum of first}}\\ \mathbf{\text{pq terms is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\left(\text{pq+1}\right)\mathbf{\text{, where p}}\mathbf{\ne }\mathbf{\text{q.}}\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Let}\text{first term of A.P.}=\mathrm{a}\\ \mathrm{Let}\text{common difference of A.P.}=\mathrm{d}\\ \mathrm{Since},{\text{​ T}}_{\text{p}}=\frac{1}{\mathrm{q}}\end{array}$

$\begin{array}{l}⇒\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}=\frac{1}{\mathrm{q}}...\left(\mathrm{i}\right)\\ \mathrm{And},{\text{​ T}}_{\text{q}}=\frac{1}{\mathrm{p}}\\ ⇒\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}=\frac{1}{\mathrm{p}}...\left(\mathrm{ii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{p}-1-\mathrm{q}+1\right)\mathrm{d}=\frac{1}{\mathrm{q}}-\frac{1}{\mathrm{p}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{p}-\mathrm{q}\right)\mathrm{d}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{pq}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{1}{\mathrm{pq}}\\ \mathrm{Substituting}\text{value of d in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\left(\mathrm{p}-1\right)\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\frac{\mathrm{p}}{\mathrm{pq}}-\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{1}{\mathrm{pq}}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2×\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}-1\right)\frac{1}{\mathrm{pq}}\right\}\\ \left[\because {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2×\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}-1\right)\frac{1}{\mathrm{pq}}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{pq}}{2}\left(\frac{2}{\mathrm{pq}}+1-\frac{1}{\mathrm{pq}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{pq}}{2}\left(\frac{1}{\mathrm{pq}}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(1+\mathrm{pq}\right)\\ \mathrm{Thus},\text{the sum of pq terms is}\frac{1}{2}\left(\mathrm{pq}+1\right).\end{array}$

Q.6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Ans.

$\begin{array}{l}\begin{array}{l}\mathrm{Given}\text{A.P. 25},\text{22},\text{19},\dots \\ \mathrm{First}\text{term of A.P.}=\text{25}\\ \text{Common difference of A.P.}=22-25\end{array}\\ \begin{array}{l}\text{ }=-3\\ \mathrm{Let}\text{sum of n terms of A.P.}=116\end{array}\\ \text{i}.\text{e}.,{\text{ S}}_{\text{n}}=116\\ \frac{\text{n}}{2}\left\{2×25+\left(\text{n}-1\right)\left(-3\right)\right\}=116\\ ⇒\text{ n}\left(50-3\text{n}+3\right)=232\\ \begin{array}{l}⇒\text{ }\left(53\text{n}-3{\text{n}}^{2}\right)=232\\ ⇒\text{ }3{\text{n}}^{2}-53\text{n}+232=0\end{array}\\ ⇒\text{ }3{\text{n}}^{2}-24\text{n}-29\text{n}+232=0\\ ⇒\text{ }3\text{n}\left(\text{n}-8\right)-29\left(\text{n}-8\right)=0\\ \begin{array}{l}⇒\text{ }\left(\text{n}-8\right)\left(3\text{n}-29\right)=0\\ ⇒\text{ n}=8,\frac{29}{3}\end{array}\\ \mathrm{Since},\text{n is a natural number.So,}\\ \begin{array}{l}\text{ n}=\text{8}\\ \text{Last term of A.P.}=\text{a}+\left(\text{n}-1\right)\text{d}\end{array}\\ \text{ }=25+\left(8-1\right)\left(-3\right)\\ \text{ }=25-21\\ \text{ }=4\\ \mathrm{Thus},\text{the last term of A.P. is 4.}\end{array}$

Q.7 Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Ans.

$\begin{array}{l}{\mathrm{k}}^{\mathrm{th}}\text{term of A.P.}=\text{5k+1}\\ ⇒\text{\hspace{0.17em}}{\mathrm{T}}_{\mathrm{k}}=\text{5k+1}\\ \mathrm{Substituting}\text{k}=1,2,3,...\text{respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{1}}=5\left(1\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{2}}=5\left(2\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{3}}=5\left(3\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\\ \mathrm{Common}\text{difference}={\mathrm{T}}_{2}-{\mathrm{T}}_{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11-6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\end{array}$

$\begin{array}{l}\mathrm{Sum}{\text{of n terms,S}}_{\text{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left\{2×6+\left(\mathrm{n}-1\right)5\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(12+5\mathrm{n}-5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(7+5\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right)\\ \mathrm{Thus},\text{the sum of n terms is}\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right).\end{array}$

Q.8 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Ans.

$\begin{array}{l}\mathrm{Given},\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\left(\mathrm{pn}+{\mathrm{qn}}^{2}\right)\\ \mathrm{Putting}\text{n}=1,2,3\text{\hspace{0.17em}}\mathrm{respectively},\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1}=\left(\mathrm{p}×1+\mathrm{q}×{1}^{2}\right)\\ \text{\hspace{0.17em}}=\mathrm{p}+\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{2}=\left(\mathrm{p}×2+\mathrm{q}×{2}^{2}\right)\\ \text{\hspace{0.17em}}=2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{3}=\left(\mathrm{p}×3+\mathrm{q}×{3}^{2}\right)\\ \text{\hspace{0.17em}}=3\text{\hspace{0.17em}}\mathrm{p}+9\text{\hspace{0.17em}}\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{1}={\mathrm{S}}_{2}-{\mathrm{S}}_{1}\\ \text{\hspace{0.17em}}=\left(2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\right)-\left(\mathrm{p}+\mathrm{q}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}=\mathrm{p}+3\mathrm{q}\\ {\mathrm{T}}_{2}={\mathrm{S}}_{3}-{\mathrm{S}}_{2}\\ \text{\hspace{0.17em}}=\left(3\text{\hspace{0.17em}}\mathrm{p}+9\text{\hspace{0.17em}}\mathrm{q}\right)-\left(2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\right)\\ \text{\hspace{0.17em}}=\mathrm{p}+5\text{\hspace{0.17em}}\mathrm{q}\\ \mathrm{Common}\text{​ difference,}\\ \text{d}={\mathrm{T}}_{2}-{\mathrm{T}}_{1}\\ \text{\hspace{0.17em}}=\left(\mathrm{p}+5\text{\hspace{0.17em}}\mathrm{q}\right)-\left(\mathrm{p}+3\mathrm{q}\right)\\ \text{\hspace{0.17em}}=2\mathrm{q}\\ \mathrm{Thus},\text{common difference is 2q.}\end{array}$

Q.9 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Ans.

$\begin{array}{l}\mathrm{Since},\\ \frac{\mathrm{Sum}\text{of n terms of an A.P.}}{\mathrm{Sum}\text{of n terms of another A.P.}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{A}+\left(\mathrm{n}-1\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left\{\mathrm{a}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{d}\right\}}{\left\{\mathrm{A}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Since},{\text{​ T}}_{\text{18}}=\text{a}+\text{17d. So, Substituting}\frac{\mathrm{n}-1}{2}=17\end{array}$

$\begin{array}{l}\mathrm{or}\text{\hspace{0.17em}}\mathrm{n}=35\text{in equation}\left(\mathrm{i}\right),\text{​​ we get}\\ \frac{\left\{\mathrm{a}+17\mathrm{d}\right\}}{\left\{\mathrm{A}+17\mathrm{D}\right\}}=\frac{\text{5}×\text{35}+\text{4}}{\text{9}×\text{35}+\text{6}}\\ \frac{{\text{18}}^{\text{th}}\text{term of an A.P.}}{{\text{18}}^{\text{th}}\text{term of another A.P.}}=\frac{\text{175}+\text{4}}{\text{315}+\text{6}}\\ =\frac{179}{321}\\ \mathrm{Thus},{\text{the ratio of 18}}^{\text{th}}\text{terms of two A.P. is 179:321.}\end{array}$

Q.10 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Ans.

$\begin{array}{l}\text{first term of A.P.}=\text{a}\\ \mathrm{and}\text{common difference of A.P.}=\text{d}\\ \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{p}}={\mathrm{S}}_{\mathrm{q}}\left(\mathrm{Given}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{2}\left\{2\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}=\frac{\mathrm{q}}{2}\left\{2\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{ap}+\mathrm{p}\left(\mathrm{p}-1\right)\mathrm{d}=2\mathrm{aq}+\mathrm{q}\left(\mathrm{q}-1\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{ap}-2\mathrm{aq}=\mathrm{q}\left(\mathrm{q}-1\right)\mathrm{d}-\mathrm{p}\left(\mathrm{p}-1\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left({\mathrm{q}}^{2}-\mathrm{q}-{\mathrm{p}}^{2}+\mathrm{p}\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left\{\mathrm{p}-\mathrm{q}-\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left\{\left(\mathrm{p}-\mathrm{q}\right)-\left(\mathrm{p}-\mathrm{q}\right)\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left(\mathrm{p}-\mathrm{q}\right)\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}=\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\end{array}$

$\begin{array}{l}\mathrm{Now},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left\{2\mathrm{a}+\left(\mathrm{p}+\mathrm{q}-1\right)\mathrm{d}\right\}\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\right\}\mathrm{d}+\left(\mathrm{p}+\mathrm{q}-1\right)\mathrm{d}\right]\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left\{1-\left(\mathrm{p}+\mathrm{q}\right)+\left(\mathrm{p}+\mathrm{q}\right)-1\right\}\mathrm{d}\\ ⇒{\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=0\\ \mathrm{Thus},\text{the sum of}\left(\mathrm{p}+\mathrm{q}\right)\text{terms is zero.}\end{array}$

Q.11

$\begin{array}{l}\text{Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.}\\ \text{Prove that}\frac{\text{a}}{\text{p}}\left(\text{q}-\text{r}\right)\text{+}\frac{\text{b}}{\text{q}}\left(\text{r}-\text{p}\right)\text{+}\frac{\text{c}}{\text{r}}\left(\text{p}-\text{q}\right)\text{ = 0.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be A and common difference be d.}\\ \mathrm{Given}:\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{p}}=\mathrm{a}⇒\mathrm{a}=\frac{\mathrm{p}}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{a}}{\mathrm{p}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\\ {\mathrm{S}}_{\mathrm{q}}=\mathrm{b}⇒\mathrm{b}=\frac{\mathrm{p}}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{b}}{\mathrm{q}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ \text{and}{\mathrm{S}}_{\mathrm{r}}=\mathrm{c}⇒\mathrm{c}=\frac{\mathrm{r}}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{c}}{\mathrm{r}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}}{\mathrm{p}}\left(\mathrm{q}-\mathrm{r}\right)+\frac{\mathrm{b}}{\mathrm{q}}\left(\mathrm{r}-\mathrm{p}\right)+\frac{\mathrm{c}}{\mathrm{r}}\left(\mathrm{p}-\mathrm{q}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\left(\mathrm{q}-\mathrm{r}\right)+\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\left(\mathrm{r}-\mathrm{p}\right)\\ +\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\left(\mathrm{p}-\mathrm{q}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\begin{array}{l}2\mathrm{A}\left(\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}\right)+\\ \mathrm{d}\left\{\left(\mathrm{p}-1\right)\left(\mathrm{q}-\mathrm{r}\right)+\left(\mathrm{q}-1\right)\left(\mathrm{r}-\mathrm{p}\right)+\left(\mathrm{r}-1\right)\left(\mathrm{p}-\mathrm{q}\right)\right\}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{2\mathrm{A}\left(0\right)+\mathrm{d}\left(\begin{array}{l}\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}\\ +\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}\end{array}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(0+0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.12 The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).

Ans.

$\begin{array}{l}\mathrm{Let}\text{​ first term of an A.P.}=\text{a}\\ \text{and commong difference of A.P.}=\mathrm{d}\\ \mathrm{It}\text{is given that:}\\ {\text{S}}_{\text{m}}{\text{:S}}_{\text{n}}={\mathrm{m}}^{2}:{\mathrm{n}}^{2}\\ ⇒\text{\hspace{0.17em}}\frac{{\mathrm{S}}_{\mathrm{m}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\end{array}$

$\begin{array}{l}⇒\frac{\frac{\mathrm{m}}{2}\left\{2\mathrm{a}+\left(\mathrm{m}-1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\\ ⇒\frac{\left\{2\mathrm{a}+\left(\mathrm{m}-1\right)\mathrm{d}\right\}}{\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}=\frac{\mathrm{m}}{\mathrm{n}}\\ ⇒2\mathrm{an}+\left(\mathrm{mn}-\mathrm{n}\right)\mathrm{d}=2\mathrm{am}+\left(\mathrm{mn}-\mathrm{m}\right)\mathrm{d}\\ ⇒2\mathrm{an}-2\mathrm{am}=\left(\mathrm{mn}-\mathrm{m}\right)\mathrm{d}-\left(\mathrm{mn}-\mathrm{n}\right)\mathrm{d}\\ ⇒\text{}2\mathrm{a}\left(\mathrm{n}-\mathrm{m}\right)=\mathrm{d}\left(\mathrm{mn}-\mathrm{m}-\mathrm{mn}+\mathrm{n}\right)\\ ⇒\text{}2\mathrm{a}\left(\mathrm{n}-\mathrm{m}\right)=\mathrm{d}\left(-\mathrm{m}+\mathrm{n}\right)\\ ⇒\text{}2\mathrm{a}=\mathrm{d}\\ \mathrm{Now},\\ \frac{{\mathrm{T}}_{\mathrm{m}}}{{\mathrm{T}}_{\mathrm{n}}}=\frac{\mathrm{a}+\left(\mathrm{m}-1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}-1\right)2\mathrm{a}}\\ =\frac{\mathrm{a}+\left(\mathrm{m}-1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}-1\right)2\mathrm{a}}\\ =\frac{1+2\mathrm{m}-2}{1+2\mathrm{n}-2}\\ =\frac{2\mathrm{m}-1}{2\mathrm{n}-1}\\ \mathrm{Thus},{\text{the ratio of m}}^{\text{th}}{\text{and n}}^{\text{th}}\text{term is}\left(2\mathrm{m}-1\right):\left(2\mathrm{n}-1\right).\end{array}$

Q.13 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Ans.

$\begin{array}{l}\mathrm{It}\text{ is given that}\\ {\text{S}}_{\text{n}}={\text{3n}}^{\text{2}}+\text{5n}\\ \text{Replace n by n}-\text{1, we get}\\ {\text{S}}_{\text{n-1}}=\text{3}{\left(\text{n}-\text{1}\right)}^{\text{2}}+\text{5}\left(\text{n}-1\right)\\ \text{ }={\text{3n}}^{\text{2}}-6\text{n}+\text{3}+\text{5n}-5\\ \text{ }={\text{3n}}^{\text{2}}-\text{n}-2\\ \mathrm{So},{\text{ T}}_{\text{n}}={\text{S}}_{\text{n}}-{\text{S}}_{\text{n-1}}\\ \text{ }=\left({\text{3n}}^{\text{2}}+\text{5n}\right)-\left({\text{3n}}^{\text{2}}-\text{n}-2\right)\\ \text{ }=6\text{n}+2\\ \mathrm{Since},{\text{ T}}_{\text{m}}=164\\ \mathrm{So},\text{ 6m}+\text{2}=\text{164}\\ ⇒\text{ m}=\frac{164-2}{6}\\ \text{ }=\frac{162}{6}\\ \text{ }=27\\ \mathrm{Thus},\text{the value of m is 27.}\end{array}$

Q.14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Ans.

$\begin{array}{l}\mathrm{Let}{\text{5 numbers between 8 and 26 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{and A}}_{\text{5}}\text{.}\\ {\text{Then, A.P. is 8, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{, A}}_{\text{5}}\text{,\hspace{0.17em}26.}\\ \because \mathrm{Last}\text{term}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}26=8+\left(7-1\right)\mathrm{d}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}26=8+6\text{\hspace{0.17em}}\mathrm{d}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{26-8}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{18}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \mathrm{Then},\text{\hspace{0.17em}}{\mathrm{A}}_{1}=8+\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{2}=8+2\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+2\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}=8+3\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+3\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{4}=8+4\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+4\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}}{\mathrm{A}}_{5}=8+5\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+5\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=23\\ \mathrm{Thus},\text{the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.}\end{array}$

Q.15

$\mathbf{\text{If}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}is the A.M. between a and b, then find the value of n.}}$

Ans.

$\begin{array}{l}\mathrm{A}.\mathrm{M}.\text{of a and b}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \text{\hspace{0.17em}}\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ ⇒\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}\right)\text{\hspace{0.17em}}=2\left({\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\right)\\ ⇒{\mathrm{a}}^{\mathrm{n}}+{\mathrm{ab}}^{\mathrm{n}-1}+{\mathrm{ba}}^{\mathrm{n}-1}+{\mathrm{b}}^{\mathrm{n}}=2{\mathrm{a}}^{\mathrm{n}}+2{\mathrm{b}}^{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{ab}}^{\mathrm{n}-1}-{\mathrm{b}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}-{\mathrm{ba}}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)=\text{\hspace{0.17em}}{\mathrm{a}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}-1}=\text{\hspace{0.17em}}{\mathrm{a}}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{n}-1}=1=\text{\hspace{0.17em}}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=1\end{array}$

Q.16 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.

Ans.

$\begin{array}{l}\mathrm{Let}{\text{m numbers between 1 and 31 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{.}\\ {\text{Then, A.P. is 1, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{,\hspace{0.17em}\hspace{0.17em}31.}\\ \because \mathrm{Last}\text{term}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}31=1+\left(\mathrm{m}+2-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}30=\left(\mathrm{m}+1\right)\text{\hspace{0.17em}}\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{30}{\mathrm{m}+1}\\ \mathrm{Then},\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{1}=1+\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+\frac{30}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{m}+31}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{7}=1+7\mathrm{d}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

$\begin{array}{l}=1+7\left(\frac{30}{\mathrm{m}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{m}+211}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{\mathrm{m}-1}=1+\left(\mathrm{m}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+\left(\mathrm{m}-1\right)\left(\frac{30}{\mathrm{m}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{m}+1+30\left(\mathrm{m}-1\right)}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{31\mathrm{m}-29}{\mathrm{m}+1}\\ \mathrm{According}\text{to question:}\\ \frac{{\mathrm{A}}_{7}}{{\mathrm{A}}_{\mathrm{m}-1}}=\frac{\left(\frac{\mathrm{m}+211}{\mathrm{m}+1}\right)}{\left(\frac{31\text{\hspace{0.17em}}\mathrm{m}-29}{\mathrm{m}+1}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5}{9}=\frac{\text{\hspace{0.17em}}\mathrm{m}+211}{31\text{\hspace{0.17em}}\mathrm{m}-29}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}155\text{\hspace{0.17em}}\mathrm{m}-145=9\text{\hspace{0.17em}}\mathrm{m}+1899\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}155\text{\hspace{0.17em}}\mathrm{m}-9\mathrm{m}=1899+145\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}146\text{\hspace{0.17em}}\mathrm{m}=2044\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{m}=\frac{2044}{146}\\ =14\\ \mathrm{Thus},\text{the value of m is 14.}\end{array}$

Q.17 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Ans.

$\begin{array}{l}\mathrm{The}\text{difference between any two consecutive}\\ \text{angles}=5\mathrm{°}\\ \mathrm{The}\text{smallest angle of a polygon}=120°\\ \mathrm{Let}\text{number of sides in given polygon be n.Then,}\\ \text{Sum of angles of a polygon}=\left(\mathrm{n}-2\right)180\mathrm{°}\\ \mathrm{The}\text{series formed by angles of polygon is:}\\ \text{120°,125°,130°,135°,140°,\hspace{0.17em}}...\end{array}$

$\begin{array}{l}\mathrm{Sum}\text{of n angles of polygon of n sides}=\frac{\mathrm{n}}{2}\left\{2×120\mathrm{°}+\left(\mathrm{n}-1\right)×5\mathrm{°}\right\}\\ =\frac{\mathrm{n}}{2}\left\{240\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}-5\mathrm{°}\right\}\\ =\frac{\mathrm{n}}{2}\left\{235\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}\right\}\\ \mathrm{Therefore},\\ \frac{\mathrm{n}}{2}\left\{235\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}\right\}=\left(\mathrm{n}-2\right)180\mathrm{°}\\ ⇒5\mathrm{°}\text{\hspace{0.17em}}{\mathrm{n}}^{2}+235\mathrm{°}\mathrm{n}=360\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}-720\mathrm{°}\\ ⇒\text{\hspace{0.17em}}5\mathrm{°}\text{\hspace{0.17em}}{\mathrm{n}}^{2}-125\mathrm{°}\mathrm{n}+720\mathrm{°}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{n}}^{2}-25\mathrm{°}\mathrm{n}+144\mathrm{°}=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{n}-9\right)\left(\mathrm{n}-16\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=9,16\\ \mathrm{Number}\text{of sides in polygon is 9 because number of angles in}\\ \text{a polygon is equal to the number of sides. Sum of 16 angles}\\ \text{will be equal to sum of 9 angles if some angles are 0° or}\\ \text{negative,}\mathrm{which}\text{​ is impossible. Therefore number of sides in}\\ \text{the polygon is 9.}\end{array}$

Q.18

${\mathbf{\text{Find the 20}}}^{\mathbf{\text{th}}}{\mathbf{\text{and n}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of the G.P.}}\frac{\mathbf{\text{5}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{4}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{8}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}$

Ans.

$The\text{given G}\text{.P}\text{. is:}\text{}\text{}\text{}\frac{5}{2},\frac{5}{4},\frac{5}{8},\dots$ $\begin{array}{l}\mathrm{a}=\frac{5}{2}\\ \mathrm{r}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)}=\frac{1}{2}\\ {\mathrm{T}}_{20}={\mathrm{ar}}^{20-1}\left[\because {\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\right]\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{19}\\ =\frac{5}{{2}^{20}}\\ \mathrm{and}\text{\hspace{0.17em}}{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{\mathrm{n}-1}\\ =\frac{5}{{2}^{\mathrm{n}}}\end{array}$

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