NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.1

NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.1

NCERT Solutions Class 11th Maths Chapter – 9 Sequences and Series Exercise 9.1

?Chapter – 9?

✍Sequences and Series✍

?Exercise 9.1?

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) 

?‍♂️solution – Given,
nth 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

?‍♂️solution – Given n^th term, a_n = 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. a_n = 2ⁿ

?‍♂️solution – Given 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

?‍♂️solution – Given n^th term, a_n = (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. a_n = (-1)^n-1 5ⁿ⁺¹

?‍♂️solution – Given n^th term, a_n = (-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 n^th terms are:

7. a_n = 4n – 3; a₁₇, a₂₄

?‍♂️solution – Given,
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

?‍♂️solution – Given,
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

?‍♂️solution – Given,
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.

?‍♂️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, a_n = 3a_n-1 + 2 for all n > 1

?‍♂️solution – 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

?‍♂️solution – Given,
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

?‍♂️solution – Given,
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 

?‍♂️solution – Given,
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,