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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

The NCERT Solutions of the first exercise of Class 11 Chapter 9 are available here. These solutions can be downloaded in PDF format as well. Chapter 9 Sequences and Series of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. Exercise 9.1 of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series is based on the following topics:

1. Introduction to Sequences and Series
2. Sequences
3. Series

The NCERT textbook contains numerous questions intended for the students to solve and practise. To obtain high marks in the Class 11 examination, solving and practising the NCERT Solutions for Class 11 Maths is a must.

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

Download PDF

carouselExampleControls112

Previous





Next

Access Other Exercise Solutions for Class 11 Maths Chapter 9 – Sequences and Series

Exercise 9.2 Solutions 18 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise on Chapter 9 Solutions 32 Questions

Also explore – NCERT Class 11 Solutions

Access Solutions for Class 11 Maths Chapter 9 Exercise 9.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. 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. a_n = n/n+1

Solution:

Given n^th term, a_n = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

3. a_n = 2ⁿ

Solution:

Given n^th term, a_n = 2ⁿ

On substituting n = 1, 2, 3, 4, 5, we get

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

4.  a_n = (2n – 3)/6

Solution:

Given n^th term, a_n = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

5. a_n = (-1)^n-1 5ⁿ⁺¹

Solution:

Given n^th term, a_n = (-1)^n-1 5ⁿ⁺¹

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, –125, 625, –3125, and 15625.

6.

Solution:

On substituting n = 1, 2, 3, 4, 5, we get first 5 terms

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n^th terms are:

7. 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

8. 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

9. a_n = (-1)^n-1 n³; a₉

Solution:

Given,

n^th term of the sequence is a_n = (-1)^n-1 n³

On substituting n = 9, we get

a₉ = (-1)^9-1 (9)³ = 1 x 729 = 729

10.

Solution:

On substituting n = 20, we get

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) + ……

14. The Fibonacci sequence is defined by

1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2}, n > 2

Find a_n+1/a_n, for n = 1, 2, 3, 4, 5 

Solution:

Given,

1 = a₁ = a₂

a_n = a_{n – 1} + a_{n – 2}, n > 2

So,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Thus,