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.

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

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

Answer

an = 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:
an = n/(n + 1)

Answer

an = 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 nth term is: an = 2n

Answer

an = 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: an = (2n – 3)/6.

Answer

Here an = (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 an = (–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:
an = (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:
an = 4n – 3, a17, a24

Answer

an = 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:
an = n2/2n ; a7

Answer

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

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

Answer

an = (–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 an = (n(n -2))/(n + 3) ; a20

Answer

an = (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, an = 3, an – 1 + 2 for all n > 1.

Answer

a1 = 3, an = 3 an –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, an = (an -1)/n , n ≥ 2

Answer

Here a1 = -1, an = (an – 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, an = an– 1 – 1, n > 2.

Answer

Given: a1 = a2 = 2
and an = an– 1 – 1, n > 2.
For n = 3, a3 = a2 – 1 = 2 – 1 = 1
For n = 4, a4 = a3 – 1 = 1 – 1 = 0
For n = 5, a5 = a4 – 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 = an – 1 + an – 2, n > 2
Find (an+ 1)/an , for n = 1, 2, 3, 4, 5

Answer

For n = 1, (an+1)/an = a2/a1 = 1/1 = 1 a1 = a2 = 1
and an = an – 1 + an – 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, (an+1)/an = a3/a2 = 2/1 = 2;
for n = 3, (an+1)/an = a4/a3 = 3/2;
for n = 4, (an+1)/an = a5/a4 = 5/3;
for n = 5, (an+1)/an = a6/a5 = 8/5
The values of (an+1)/an = for n = 1, 2, 3, 4, 5 are 1, 2, 3/2 , 5/3 , 8/5
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