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

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. The 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

NCERT textbook contains a lot of questions intended for the students to solve and practice. To obtain high marks in the Class 11 examination, solving and practicing the NCERT Solutions for Class 11 Maths is a must.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.1     ### Access Other Exercise Solutions of 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

#### 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 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 = 1, 2, 3, 4, 5, we get 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 = 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 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 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, 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, 