## 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:
an = n (n + 2)

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)

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

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.

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

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

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

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

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

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

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.

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

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.

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

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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