NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1

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

a₁ = 1(1 + 2) = 3

a₂ = 2(2 + 2) = 8

a₃ = 3(3 + 2) = 15

a₄ = 4(4 + 2) = 24

a₅ = 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; a₁₇, a₂₄

Solution:

Given,

n^th term of the sequence is a_n = 4n – 3

On substituting n = 17, we get

a₁₇ = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a₂₄ = 4(24) – 3 = 96 – 3 = 93

2. a_n = n²/2ⁿ ; a⁷

Solution:

Given,

n^th term of the sequence is a_n = n²/2ⁿ

Now, on substituting n = 7, we get

a₇ = 7²/2⁷ = 49/ 128

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

11. a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1

Solution:

Given, a_n = 3a_n-1 + 2 and a₁ = 3

Then,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 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. a₁ = -1, a_n = a_n-1/n, n ≥ 2

Solution:

Given,

a_n = a_n-1/n and a₁ = -1

Then,

a₂ = a₁/2 = -1/2

a₃ = a₂/3 = -1/6

a₄ = a₃/4 = -1/24

a₅ = a₄/5 = -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. a₁ = a₂ = 2, a_n = a_n-1 – 1, n > 2

Solution:

Given,

a₁ = a₂, a_n = a_n-1 – 1

Then,

a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 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) + ……