# 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. Chapter 9 Sequences and Series of Class 11 Maths is categorized under the term â€“ I CBSE Syllabus for 2021-22. 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 term â€“ I examination, solving and practicing the NCERT Solutions for Class 11 Maths is a must.

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### 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)Â

Solution:

Given,

nth term of a sequence an = n (n + 2)Â

On substitutingÂ nÂ = 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.Â an = n/n+1

Solution:

Given nth term, an = n/n+1

On substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

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

3. anÂ = 2n

Solution:

Given nth term, anÂ = 2n

On substitutingÂ nÂ = 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. Â anÂ = (2n â€“ 3)/6

Solution:

Given nth term, anÂ = (2n â€“ 3)/6

On substitutingÂ nÂ = 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 substitutingÂ nÂ = 1, 2, 3, 4, 5, we get

Hence, the required terms are 25, â€“125, 625, â€“3125, and 15625.

6.

Solution:

On substitutingÂ nÂ = 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 â€“ 3; a17, a24

Solution:

Given,

nthÂ term of the sequence is an = 4n â€“ 3

On substitutingÂ nÂ = 17, we get

a17 = 4(17) â€“ 3 = 68 â€“ 3 = 65

Next, on substitutingÂ nÂ = 24, we get

a24 = 4(24) â€“ 3 = 96 â€“ 3 = 93

8. anÂ =Â n2/2nÂ ;Â a7

Solution:

Given,

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

Now, on substitutingÂ nÂ = 7, we get

a7 = 72/27 = 49/ 128

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

Solution:

Given,

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

On substitutingÂ nÂ = 9, we get

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

10.

Solution:

On substitutingÂ nÂ = 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 â€¦â€¦.

12. a1 = -1, an = an-1/n, n â‰¥ 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) + â€¦â€¦.

13. a1 = a2 = 2, an = an-1 â€“ 1, n > 2

Solution:

Given,

a1 = a2, an = an-1 â€“ 1

Then,

a3 = a2 â€“ 1 = 2 â€“ 1 = 1

a4 = a3 â€“ 1 = 1 â€“ 1 = 0

a5 = a4 â€“ 1 = 0 â€“ 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) + â€¦â€¦

14. The Fibonacci sequence is defined by

1 = a1 = a2 and an = an â€“ 1 + an â€“ 2, n > 2

Find an+1/an, for n = 1, 2, 3, 4, 5Â

Solution:

Given,

1 = a1 = a2

an = an â€“ 1 + an â€“ 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,

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