NCERT Solutions for Exercise 8.1, Class 11, Maths

Exercise 8.1
Sequence and Series
Question and Answers
Class 11 – Maths

Class	Class 11
Subject	Mathematics
Chapter Name	Sequences and Series
Chapter No.	Chapter 8
Exercise	Exercise 8.1
Category	Class 11 Maths NCERT Solutions

1. Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are :

(1) a_n = n (n + 2)

Answer n^th term of a sequence a_n = 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)  a_n = n/n+1

Answer Given the n^th term, a_n = n/n+1

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

a₁ = 1 / (1 + 1) = 1/2
a₂ = 2 / (2 + 1) = 2/3
a₃ = 3 / (3 + 1) = 3/4
a₄ = 4 / (4 + 1) = 4/5
a₅ = 5 / (5 + 1) = 5/6

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

(3) a_n = 2ⁿ

Answer Given the n^th term, a_n = 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) a_n = (2n – 3)/6

Answer Given the n^th term, a_n = (2n – 3)/6

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

a₁ = (2(1) – 3) / 6 = -1/6
a₂ = (2(2) – 3) / 6 = 1/6
a₃ = (2(3) – 3) / 6 = 3/6 = 1/2
a₄ = (2(4) – 3) / 6 = 5/6
a₅ = (2(5) – 3) / 6 = 7/6

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

(5) a_n = (-1)^n-1 5ⁿ⁺¹

Answer Given the n^th term, a_n = (-1)^n-1 5ⁿ⁺¹

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

a₁ = (-1)^1-1 5¹⁺¹ = (-1)⁰ 5² = 25
a₂ = (-1)^2-1 5²⁺¹ = (-1)¹ 5³ = -125
a₃ = (-1)^3-1 5³⁺¹ = (-1)² 5⁴ = 625
a₄ = (-1)^4-1 5⁴⁺¹ = (-1)³ 5⁵ = -3125
a₅ = (-1)^5-1 5⁵⁺¹ = (-1)⁴ 5⁶ = 15625

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

(6) a_n = n (n² + 5) / 4

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

a₁ = 1 · (1² + 5) / 4 = 6/4 = 3/2
a₂ = 2 · (2² + 5) / 4 = 2 · 9/4 = 9/2
a₃ = 3 · (3² + 5) / 4 = 3 · 14/4 = 21/2
a₄ = 4 · (4² + 5) / 4 = 4 · 21/4 = 21
a₅ = 5 · (5² + 5) / 4 = 5 · 30/4 = 75/2

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 n^th terms are:

(7) a_n = 4n – 3; a₁₇, a₂₄

Answer The n^th term of the sequence is a_n = 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) a_n = n²/2ⁿ ; a⁷

Answer The n^th term of the sequence is a_n = n²/2ⁿ

Now, on substituting n = 7, we get

a₇ = 7²/2⁷ = 49/ 128

(9) a_n = (-1)^n-1 n³; a₉

Answer The n^th term of the sequence is a_n = (-1)^n-1 n³

On substituting n = 9, we get

a₉ = (-1)^9-1 (9)³ = 1 x 729 = 729

(10) a_n = n (n² + 5 ) /4  ; a₂₀

Answer On substituting n = 20, we get

a₂₀ = 20(20 – 2) / (20 + 3) = (20 × 18) / 23 = 360/23

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

(11) a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1

Answer Given, a_n = 3a_n-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, a_n = a_n-1/n, n ≥ 2

Answer a_n = a_n-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, a_n = a_n-1 – 1, n > 2

Answer a₁ = a₂, a_n = a_n-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 a_n= a_{n – 1} + a_{n – 2}, n >2 Find a_n+1/a_n, for n = 1, 2, 3, 4, 5 

Answer 1 = a₁ = a₂

a_n = a_{n – 1} + a_{n – 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,

For n = 1: a₂ / a₁ = 1 / 1 = 1
For n = 2: a₃ / a₂ = 2 / 1 = 2
For n = 3: a₄ / a₃ = 3 / 2
For n = 4: a₅ / a₄ = 5 / 3
For n = 5: a₆ / a₅ = 8 / 5