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**NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2-** In the previous exercise, you have already learned about the sequence, and series. The progression is a sequence of a specific pattern that you will learn in the NCERT solutions for Class 11 Maths chapter 9 exercise 9.2. The sequence is called an arithmetic progression if the difference between two consecutive terms is constant. In NCERT book exercise 9.2 Class 11 Maths, you will learn about the arithmetic progression(A. P.), properties of arithmetic progression, the sum of the arithmetic progression, arithmetic mean, etc. Also, you will get questions related to finding the nth term of arithmatic progression, using properties of arithmetic progression to asthmatic mean, etc.

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- NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2- Download Free PDF
- Download the PDF of NCERT Solutions for Class 11 Maths Chapter 9: Sequences and Series Exercise 9.2
- Access Sequences And Series Class 11 Chapter 9 Exercise: 9.2
- Question:1 Find the sum of odd integers from 1 to 2001.
- More About NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-
- Benefits of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2:-
- Key Features of NCERT 11th Class Maths Exercise 9.2 Answers
- NCERT Solutions of Class 11 Subject Wise
- Subject Wise NCERT Exampler Solutions

As you have already studied the arithmetic progression, nth term of the arithmetic progression, the sum of the arithmetic progression in the previous classes you won't get any difficulty to understand the Class 11 Maths chapter 9 exercise 9.2. This exercise is very important to solving some real-life problems as well as understanding upcoming exercises of this chapter. NCERT syllabus Exercise 9.2 Class 11 Maths is the basic exercise to understand the arithmatic progression and upcoming concepts like arithmetic mean, geometric mean, the relation between the arithmetic mean and geometric mean etc. Check NCERT Solutions if you are looking for NCERT solutions for Science and Maths of other Classes as well. You will get detailed solutions from Classes 6 to 12 for Science and Maths.

**Also, see**

- Sequences and Series Exercise 9.1
- Sequences and Series Exercise 9.3
- Sequences and Series Exercise 9.4
- Sequences and Series Miscellaneous Exercise

****As per the CBSE Syllabus for 2023-24, this chapter has been renumbered as Chapter 8.**

Answer:

Odd integers from 1 to 2001 are

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know ,

The , sum of odd integers from 1 to 2001 is 1002001.

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

Answer:

Numbers divisible by 5 from 100 to 1000 are

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know ,

The sum of numbers divisible by 5 from 100 to 1000 is 98450.

Answer:

First term =a=2

Let the series be

Sum of first five terms

Sum of next five terms

Given : The sum of the first five terms is one-fourth of the next five terms.

To prove :

L.H.S :

Hence, 20th term is –112.

Question:4 How many terms of the A.P. are needed to give the sum –25?

Answer:

Given : A.P. =

Given : sum = -25

Answer:

Given : In an A.P., if pth term is 1/q and qth term is 1/p

Subtracting (2) from (1), we get

Putting value of d in equation (1),we get

Hence,the sum of first pq terms is 1/2 (pq +1), where .

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

Answer:

Given : A.P. 25, 22, 19, ….....

a=25 , d = -3

n could not be so n=8.

Last term

The, last term of A.P. is 4.

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

Answer:

Given :

class 11 maths ch 9 ex 9.Comparing LHS and RHS , we have

and

Putting value of d,

Question:8 If the sum of n terms of an A.P. is , where p and q are constants, find the common difference

Answer:

If the sum of n terms of an A.P. is ,

Comparing coefficients of on both side , we get

The common difference of AP is 2q.

Answer:

Given: The sums of n terms of two arithmetic progressions are in the ratio.

There are two AP's with first terms = and common difference =

Substituting n=35,we get

Thus, the ratio of the 18th term of AP's is

Answer:

Let first term of AP = a and common difference = d.

Then,

Given :

Now,

Thus, sum of p+q terms of AP is 0.

Question:11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that

Answer:

To prove :

Let and d be the first term and the common difference of AP, respectively.

According to the given information, we have

Subtracting equation (2) from (1), we have

Subtracting equation (3) from (2), we have

Equating values of d, we have

Dividing both sides from pqr, we get

Hence, the given result is proved.

Question:12 The ratio of the sums of m and n terms of an A.P. is . Show that the ratio of mth and nth term is .

Answer:

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is .

To prove : the ratio of mth and nth term is .

Put , we get

From equation (1) ,we get

Hence proved.

Question:13 If the sum of n terms of an A.P. is and its term is 164, find the value of m.

Answer:

Given : If the sum of n terms of an A.P. is and its term is 164

Let a and d be first term and common difference of a AP ,respectively.

Sum of n terms =

Comparing the coefficients of n on both side , we have

Also ,

m th term is 164.

Hence, the value of m is 27.

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

Answer:

Let five numbers be A,B,C,D,E.

Then

Here we have,

Thus, we have

Thus, the five numbers are 11,14,17,20,23.

Question:15 If is the A.M. between a and b, then find the value of n.

Answer:

Given : is the A.M. between a and b.

Thus, value of n is 1.

Answer:

Let A,B,C.........M be m numbers.

Then,

Here we have,

Given : the ratio of and numbers is 5 : 9.

Putting value of d from above,

Thus, value of m is 14.

Answer:

The first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

We have ,

Thus, he will pay Rs. 245 in the 30th instalment.

Answer:

The angles of polygon forms AP with common difference of and first term as .

We know that sum of angles of polygon with n sides is

Sides of polygon are 9 or 16.

Class 11 Maths chapter 9 exercise 9.2 consists of questions related to finding the general terms of arithmetic progression, the sum of the n terms of arithmetic progression, arithmetic mean of the progression, etc. There are some properties of arithmetic progression like adding and subtracting the constant from each term of the progression, multiplying and dividing by the non zero constant to each term of the progression, etc given before the Class 11 Maths chapter 9 exercise 9.2. There are five solved examples given before this exercise. You must go through these properties of arithmetic progression and solved examples before solving exercise 9.2 Class 11 Maths.

**Also Read| **Sequences And Series Class 11 Notes

- NCERT solutions for Class 11 Maths chapter 9 exercise 9.2 are designed by the subject matter experts based on the guideline given by CBSE
- As you are familiar with arithmetic progression, it won't take much effort to get command on Class 11 Maths chapter 9 exercise 9.2
- You can use Class 11 Maths chapter 9 exercise 9.2 solutions as revision notes to quickly revise the important concepts.

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Download EBook**Comprehensive Scope:**These ex 9.2 class 11 answers address all the exercises 9.2 problems of the NCERT 11th Class Mathematics textbook.**Step-by-Step Approach:**The solutions are presented in a step-by-step manner, making it easier for students to understand and follow the problem-solving process.**Clarity and Precision:**The class 11 ex 9.2 answers are written with utmost clarity and precision, ensuring that students can grasp the mathematical concepts and techniques required for effective problem-solving.**Appropriate Mathematical Notation:**These 11th class maths exercise 9.2 answers use the correct mathematical notations and terminology, aiding students in becoming proficient in the language of mathematics.**Free Accessibility:**Typically, these class 11 maths chapter 9 exercise 9.2 solution are available free of charge, allowing students to access them without any cost, providing a valuable resource for self-study.**Supplementary Learning Aid:**These ex 9.2 class 11 answers can serve as a supplementary learning resource to reinforce classroom teachings and assist students in preparing for exams.**Homework and Practice:**Students can use these answers to check their work, practice problem-solving, and enhance their overall performance in mathematics.

**Also see-**

**Happy learning!!!**

1. If a constant term is added to each term of an A.P. the resulting sequence is ?

When a constant term is added to each term of an A.P. the resulting sequence is also an A.P.

2. If a constant term is subtracted from the each term of an A.P. the resulting sequence is ?

When a constant term is subtracted from each term of an A.P. the resulting sequence is also an A.P.

3. Find the 5th term of A.P. 20, 18, 16, ...?

Given a= 20

d = 18 -20 = -2

a_5 = a + 4d

a_5 = 20 -2(4) = 20 - 8 = 12

4. Find the sum first 10 natural numbers ?

n = 10

S_n = n(n+1)/2

S_(10) = 10x11/2 = 550

5. Find the sum of first 5 term of A.P. 20, 18, 16, ...?

Given a = 20

d = 18 -20 = -2

S_n = n/2[2a + (n-1)d]

S_5 = 5/2[40 + 4x(-2)]

S_5 = 5/2[32]

S_5 = 80

6. What is the weightage of Sequences and Series in the CBSE Class 11 Maths ?https://school.careers360.com/ncert/ncert-books

CBSE doesn't provide the chapter-wise marks distributions for CBSE Class 11 Maths exam. The whole unit algebra has 30 marks weightage out of 80 marks in the CBSE 11 Maths exam.

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