# Sequences and Series Chapter 9 Class 11 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 11 Maths Chapter 9‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 11 Maths Chapter 9 Exercise Solution’ using the download button.

### Chapter 9: Sequences and Series

#### 9.1 Introduction

In mathematics, the word, “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence,

we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member, and so on.

For example, the population of human beings or bacteria at different times forms a sequence. The amount of money deposited in a bank, over a number of years forms a sequence.

Depreciated values of certain commodities occur in a sequence. Sequences have important applications in several spheres of human activities.

Sequences, following specific patterns, are called progressions. In the previous class, we have studied arithmetic progression (A.P).

In this Chapter, besides discussing more A.P.; arithmetic mean, geometric mean, the relationship between A.M.and G.M.,

special series in the forms of the sum to in terms of consecutive natural numbers sum to n in terms of squares of natural numbers, and sum to n in terms of cubes of natural numbers will also be studied.

#### 9.2 Sequences

Let us consider the following examples: Assume that there is a generation gap of 30 years, we are asked to find the number of ancestors, i.e., parents, grandparents, great grandparents, etc. that a person might have over 300 years.

### NCERT Solutions Class 11 Maths Chapter 9 Sequences and Series

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.