NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1

NCERT Solutions for Class 11 Maths Chapter 9 Sequence and Series Exercise 9.1

Chapter 9 Sequence and Series Exercise 9.1 NCERT Solutions for Class 11 Maths is very important topic which you need to improve your marks in the examinations. Here you will find Class 11 Maths NCERT Solutions that will help you a lot in understanding the basics things of the chapter and implementing them. It will save your precious time if you want to complete your homework on time.

1. Write the first five terms of each of the sequence defined by the following:
a_n = n (n + 2)

Answer

a_n = n (n + 2)
For n = 1, a₁ = 1(1 + 2) = 3
For n = 2, a₂ = 2(2 + 2) = 8
For n = 3, a₃ = 3(3 + 2) = 15
For n = 4, a₄ = 4(4 + 2) = 24
For n = 5, a₅ = 5(5 + 2) = 35
Thus first five terms are 3, 8, 15, 24, 35.

2. Write the first 5 terms of the sequence defined by the following:
a_n = n/(n + 1)

Answer

a_n = n/(n + 1)
for n = 1, a₁ = 1/(1 + 1) = 1/2
for n = 2, a₂ = 2/(2 +1 ) = 2/3
for n = 3, a₃ = 3/(3 + 1) = 3/4
for n = 4, a₄ = 4/(4 + 1) = 4/5
for n = 5, a₅ = 5/(5 + 1) = 5/6
Hence, first 5 terms of the given sequence are
1/2, 2/3, 3/4, 4/5 and 5/6.

3. Write the first five terms of the sequence whose n^th term is: a_n = 2ⁿ

Answer

a_n = 2ⁿ, Putting n = 1, 2, 3, 4, 5
a₁ = 2¹ = 2, a₂ = 2² = 4, a₃ = 2³ = 8
a ₄ = 2⁴ = 16, a₅ = 2⁵ = 32
Required first five term of sequences are 2, 4, 8, 16, 32.

4. Write the first five terms of the sequence whose n ^th term is: a_n = (2n – 3)/6.

Answer

Here a_n = (2n – 3)/6
Putting n = 1, 2, 3, 4, 5, we get
a₁ = (2 × 1 – 3)/6 = (2 – 3)/6 = -(1/6);
a₂ = (2 × 2 – 3)/6 = (4 – 3)/6 = 1/6;
a₃ = (2 × 3 – 3)/6 = (6 – 3)/6 = 3/6 = 1/2;
a₄ = (2 × 4 – 3)/6 = (8 – 3)/6 = 5/6
and a₅ = (2 × 5 – 3)/6 = (10 – 3)/6 = 7/6
∴ the first five terms are –(1/6), 1/6, 1/2, 5/6 and 7/6.

5. Write the first five terms of the sequence whose n ^th term is a_n = (–1)^{n – 1} 5 ^{n + 1} 

Answer

Putting n = 1, 2, 3, 4, 5
a₁ = (–1)⁰. 5 ^{1 + 1} = 5² = 25
a₂ = (–1)¹. 5 ^{2 + 1} = – 5³ = –125
a₃ = (–1)². 5 ^{3 + 1} = 5⁴ = 625
a₄ = (–1)³. 5 ^{4 + 1} = – 5⁵ = – 3125
a₅ = (–1)⁴. 5 ^{5 + 1} = 5⁶ = 15625

6. Write the first 5 terms of the sequence defined by the following:
a_n = (n(n² + 5))/4

Answer

for n = 1, a₁ = (1(1² + 5))/4 = 6/4 = 3/2
for n =2 , a₂ = (2(2² + 5))/4 = (2 × 9)/4 = 9/2
for n = 3, a₃ = (3(3² + 5))/4 = (3 × 14)/4 = 21/2
for n = 4, a₄ = (4(4² + 5))/4 = (4 × 21)/4 = 21
for n = 5, a₅ = (5(5² + 5))/4 = (5 × 30)/4 = 75/2
hence the first 5 terms are 3/2, 9/2, 21/2, 21 and 75/2

7. Find the indicated terms in the following sequence whose n ^th term is:
a_n = 4n – 3, a₁₇, a₂₄

Answer

a_n = 4n – 3
For n = 17, a₁₇ = 4 × 17 – 3 = 68 – 3 = 65
For n = 24, a₂₄ = 4 × 24 – 3 = 96 – 3 = 93
Hence a₁₇ = 65 and a₂₄ = 93

8. Find the indicated terms in the following sequence whose n ^th term is:
a_n = n²/2ⁿ ; a₇

Answer

a_n = n²/2ⁿ
putting n = 7
a₇ = 7²/2⁷ = 49/128

9. Find the indicated term in the following sequence whose n ^th term is:
a_n = (–1)^{n – 1} n ³, a₉

Answer

a_n = (–1)^{n – 1} n³, Putting n = 9
a₉ = (–1)^{9 – 1} 9³ = 729

10. Find the indicated terms in the sequence whose n ^th term is a_n = (n(n -2))/(n + 3) ; a₂₀

Answer

a_n = (n(n -2))/(n + 3),
Putting n = 20,
a₂₀ = (20(20 – 2))/(20 + 3) = (20 × 18)/23 = 360/23

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

a₁ = 3, a_n = 3, a_{n – 1} + 2 for all n > 1.

Answer

a₁ = 3, a_n = 3 a_{n –1} + 2. Putting n = 2, 3, 4, 5
a₂ = 3.a₁ + 2 = 3. 3 + 2 = 9 + 2 = 11
a₃ = 3.a₂ + 2 = 3. 11 + 2 = 33 + 2 = 35
a₄ = 3.a₃ + 2 = 3. 35 + 2 = 105 + 2 = 107
a₅ = 3.a₄ + 2 = 3. 107 + 2 = 321 + 2 = 323
The first five terms of the sequences are
3, 11, 35, 107, 323
∴ The corresponding series is
3 + 11 + 35 + 107 + 323 +....

12. a₁ = -1, a_n = (a_{n -1})/n , n ≥ 2

Answer

Here a₁ = -1, a_n = (a_{n – 1})/n , n ≥ 2
Putting n = 2, 3, 4, 5 and 6
a₂ = (a₂ – 1)/2 = a₁/2 = -1/2
a₃ = (a₃ – 1)/3 = a₂/2 = -(1/2)/3 = -1/6;
a₄ = (a₄ – 1)/4 = a₃/4 = -(1/6)/4 = -1/24;
a₅ = a₅-1/5 = a₄/5 = -(1/24)/5 = -1/120
The first five terms of the sequences are
-1, -(1/2), -(1/6), -(1/24), -1/120
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

Given: a₁ = a₂ = 2
and a_n = a_{n– 1} – 1, n > 2.
For n = 3, a₃ = a_{2 – 1} = 2 – 1 = 1
For n = 4, a₄ = a_{3 – 1} = 1 – 1 = 0
For n = 5, a₅ = a_{4 – 1} = 0 – 1 = –1
Hence first five terms are 2, 2, 1, 0, –1 and the corresponding series = 2 + 2 + 1 + 0 + (–1) +....

14. The Fibonacci sequence is defined by
1 = a₁ = a₂ and an = a_{n – 1} + a_{n – 2}, n > 2
Find (a_{n+ 1})/a_n , for n = 1, 2, 3, 4, 5

Answer

For n = 1, (a_n+1)/a_n = a₂/a₁ = 1/1 = 1 a₁ = a₂ = 1
and a_n = a_{n – 1} + a_{n – 2}, n > 2 …(A)
n = 3 in eqn (A) a₃ = a₂ + a₁ = 1 + 1 = 2
n = 4 in eqn (A) a₄ = a₃ + a₂ = 2 + 1 = 3
n = 5 in eqn (A) a₅ = a₄ + a₃ = 3 + 2 = 5
n = 6 in eqn (A) a₆ = a₅ + a₄ = 5 + 3 = 8
for n = 2, (a_n+1)/a_n = a₃/a₂ = 2/1 = 2;
for n = 3, (a_n+1)/a_n = a₄/a₃ = 3/2;
for n = 4, (a_n+1)/a_n = a₅/a₄ = 5/3;
for n = 5, (a_n+1)/a_n = a₆/a₅ = 8/5
The values of (a_n+1)/a_n = for n = 1, 2, 3, 4, 5 are 1, 2, 3/2 , 5/3 , 8/5