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:

- Introduction to Sequences and Series
- Sequences
- 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.

#### 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* = 2*n*

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

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

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,

*n*th 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,

*n*th 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,

*n*th 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,