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

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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. an = n (n + 2) 

Solution:

Given,

nth term of a sequence an = 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. an = n/n+1

Solution:

Given nth term, an = n/n+1

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 1

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

3. an = 2n

Solution:

Given nth term, an = 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.  an = (2n – 3)/6

Solution:

Given nth term, an = (2n – 3)/6

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 2

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

5. an = (-1)n-1 5n+1

Solution:

Given nth term, an = (-1)n-1 5n+1

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 3

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

6.

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 4

Solution:

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

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 5

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 nth terms are:

7. an = 4n – 3; a17, a24

Solution:

Given,

nth term of the sequence is an = 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. an = n2/2n ; a7

Solution:

Given,

nth term of the sequence is an = n2/2n

Now, on substituting n = 7, we get

a7 = 72/27 = 49/ 128

9. an = (-1)n-1 n3; a9

Solution:

Given,

nth term of the sequence is an = (-1)n-1 n3

On substituting n = 9, we get

a9 = (-1)9-1 (9)3 = 1 x 729 = 729

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 6

10.

Solution:

On substituting n = 20, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 7

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

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

Solution:

Given, an = 3an-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, an = an-1/n, n ≥ 2

Solution:

Given,

an = an-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, an = an-1 – 1, n > 2

Solution:

Given,

a1 = a2, an = an-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 an = an – 1 + an – 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5 

Solution:

Given,

1 = a1 = a2

an = an – 1 + an – 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,

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 8

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