Table of Contents

The **NCERT Solutions** for the first exercise of Class 11 Maths Chapter 9 are available here and can also be downloaded in PDF format. Chapter 9, Sequences and Series, is part of the **CBSE Syllabus** for 2024-25. Exercise 9.1 of NCERT Solutions for Class 11 Maths covers the following topics:

- Introduction to Sequences and Series
- Sequences
- Series

The **NCERT textbook** includes numerous questions for students to solve and practice. To achieve high marks in the Class 11 examination, it is essential to solve and practice the **NCERT Solutions for Class 11 Maths**.

## NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1

Download the free PDF of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1, prepared by expert Mathematics teachers at Infinity Learn, following CBSE (NCERT) guidelines. Access our Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 solutions to help you revise the entire syllabus and score higher marks in your exams.

### Access NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

**1. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:**** a _{n} = n (n + 2) **

**Solution:**

Given,

n^{th} term of a sequence a_{n} = n (n + 2)

On substituting *n* = 1, 2, 3, 4, and 5, we get the first five terms

a_{1} = 1(1 + 2) = 3

a_{2} = 2(2 + 2) = 8

a_{3} = 3(3 + 2) = 15

a_{4} = 4(4 + 2) = 24

a_{5} = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

**2. Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n ^{th} terms are:**

**1. a _{n} = 4n – 3; a_{17}, a_{24}**

**Solution:**

Given,

*n*^{th} term of the sequence is a_{n} = 4n – 3

On substituting *n* = 17, we get

a_{17} = 4(17) – 3 = 68 – 3 = 65

Next, on substituting *n* = 24, we get

a_{24} = 4(24) – 3 = 96 – 3 = 93

**2. a _{n} = n^{2}/2^{n} ; a^{7}**

**Solution:**

Given,

*n*^{th} term of the sequence is a_{n} = n^{2}/2^{n}

Now, on substituting *n* = 7, we get

a_{7} = 7^{2}/2^{7} = 49/ 128

**3. Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:**

**11. a _{1} = 3, a_{n} = 3a_{n-1} + 2 for all n > 1**

**Solution:**

Given, a_{n} = 3a_{n-1} + 2 and a_{1} = 3

Then,

a_{2} = 3a_{1} + 2 = 3(3) + 2 = 11

a_{3} = 3a_{2} + 2 = 3(11) + 2 = 35

a_{4} = 3a_{3} + 2 = 3(35) + 2 = 107

a_{5} = 3a_{4} + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

**12. a _{1} = -1, a_{n} = a_{n-1}/n, n ≥ 2**

**Solution:**

Given,

a_{n} = a_{n-1}/n and a_{1} = -1

Then,

a_{2} = a_{1}/2 = -1/2

a_{3} = a_{2}/3 = -1/6

a_{4} = a_{3}/4 = -1/24

a_{5} = a_{4}/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

**13. a _{1} = a_{2 }= 2, a_{n} = a_{n-1} – 1, n > 2**

**Solution:**

Given,

a_{1} = a_{2}, a_{n} = a_{n-1} – 1

Then,

a_{3} = a_{2} – 1 = 2 – 1 = 1

a_{4} = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{4} – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……