# Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.1

• Last Updated : 28 Dec, 2020

### Question 1. Write the first five terms of a sequence whose nth term is an=n(n+2).

Solution:

Given, an = n(n + 2)

Putting value of n as 1, 2, 3, 4 and 5. We get,

a1=1(1+2)=1(3)=3

a2=2(2+2)=2(4)=8

a3=3(3+2)=3(5)=15

a4=4(4+2)=4(6)=24

a5=5(5+2)=5(7)=35

Therefore, the first 5 terms of given series are 3, 8, 15, 24 and 35.

### Question 2. Write the first 5 terms of the series whose nth term is an = n/(n + 1)  .

Solution:

Given, an = n/(n+1)

Putting values of n as 1,2,3,4 and 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

Therefore, the first 5 terms of the given series are 1/2 , 2/3 , 3/4 , 4/5 and 5/6.

### Question 3. Write the first five terms of series whose nth term is an = 2n.

Solution:

Given, an=2n

Putting value of n as 1, 2, 3, 4 and 5. We get,

a1=21=2

a2=22=4

a3=23=8

a4=24=16

a5=25=32

Therefore, the first 5 terms of the given series are 2, 4, 8, 16 and 32.

### Question 4. Write the first five terms of series whose nth term is an = (2n – 3)/6

Solution:

Given, an=(2n -3)/6

Putting value of n as 1, 2, 3, 4 and 5. We get,

a1=(2(1) -3)/6=-1/6

a2=(2(2) -3)/6=1/6

a3=(2(3) -3)/6=1/2

a4=(2(4) -3)/6=5/6

a5=(2(5) -3)/6=7/6

Therefore, the first 5 terms of the given series are -1/6, 1/6, 1/2, 5/6 and 7/6.

### Question 5. Write the first 5 terms of the sequence whose nth term is an = (-1)n-15n+1.

Solution:

Given, an = (-1)n-15n+1

Putting values of n as 1, 2, 3, 4 and 5. We get,
a1 = (-1)1-151+1=(-1)052=25

a2 = (-1)2-152+1=(-1)153=-125

a3 = (-1)3-153+1=(-1)254=625

a4 = (-1)4-154+1=(-1)355=-3125

a5 = (-1)5-155+1=(-1)456=15625

Therefore, the first 5 terms of the series are 25, -125, 625, -3125 and 15625.

### Question 6. Find the first five terms of the series whose nth term is given as an = n(n2+5)/4.

Solution:

Given, an = n(n2+5)/4

Putting values of n as 1, 2, 3, 4 and 5. We get,

a1 = 1(12+5)/4=1(1+5)/4=1(6)/4=3/2

a2 = 2(22+5)/4=2(4+5)/4=2(9)/4=9/2

a3 = 3(32+5)/4=3(9+5)/4=3(14)/4=3(7)/2=21/2

a4 = 4(42+5)/4=4(16+5)/4=4(21)/4=21

a5 = 5(52+5)/4=5(25+5)/4=5(30)/4=5(15)/2=75/2

Therefore, the first 5 terms of the series are 3/2, 9/2, 21/2, 21 and 75/2.

### Question 7. Find the 17th term of the sequence whose nth term is given as an=4n – 3.

Solution:

Given, an = 4n – 3

Putting value of n as 17. We get,

a17 = 4(17) -3 = 68 – 3=65

Therefore, the 17th term is 65.

### Question 8. Find the 7th term of the sequence whose nth term is given as an = n2/(2n).

Solution:

Substituting n = 7. We get,

a7 = 72/(2*7)=49/14=7/2.

Therefore, the 7th term is7/2.

### Question 9. Find the 9th term of the sequence whose nth term is given as an = (-1)n-1n3.

Solution:

Substituting n = 9. We get,

a9 = (-1)9-193=(-1)893=729.

Therefore, the 9th term is 729.

### Question 10. Find the 20th term of the sequence whose nth term is gives as an = (n(n-2))/(n+3).

Solution:

Given, an = (n(n-2))/(n+3)

Putting value of n as 20. We get,

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

Therefore, the 20th term is 360/23.

### Question 11. Find the first 5 terms of the following sequence. a1 = 3, an = 3an-1+ 2 for all n>1.

Solution:

a1 = 3

a2 = 3a2-1+2=3a1+2=3*3+2=9+2=11

a3 = 3a3-1+2=3a2+2=3*11+2=33+2=35

a4 = 3a4-1+2=3a3+2=3*35+2=105+2=107

a5 = 3a5-1+2=3a4+2=3*107+2=321+2=323

The first 5 terms are 3, 11, 35, 107 and 323.

### Question 12. Write the first 5 terms of the following sequence. a1 = -1, an = an-1/n for all n>1.

Solution:

a1=-1

a2=a2-1/2=a1/2=-1/2

a3=a3-1/3=a2/3=(-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

Therefore, the first 5 terms of the series are -1,-1/2, -1/6, -1/24 and -1/120.

### Question 13. Write the first 5 terms of the following sequence . a1 = a2 = 2, an = an-1-1 for all n > 2.

Solution:

a1 = 2

a2 = 2

a3 = a3-1-1=a2-1=2-1=1

a4 = a4-1-1=a3-1=1-1=0

a5 = a5-1-1=a4-1=0-1=-1

Therefore, the first 5 terms of the series are 2, 2, 1, 0 and -1.

### Question 14. The Fibonacci sequence is given by, 1 = a1 = a2 and an = an-1 + an-2, n > 2. Find an+1/an for the first 5 terms.

Solution:

a1 = 1, a2 = 1

a3 = a1+a2=1+1=2

a4 = a2+a3=1+2=3

a5 = a3+a4=2+3=5

a6 = a4+a5=3+5=8

Value of an+1/an for,

n = 1 is a1+1/a1=a2/a1=1/1=1

n = 2 is a2+1/a2=a3/a2=2/1=2

n = 3 is a3+1/a3=a4/a3=3/2

n = 4 is a4+1/a4=a5/a4=5/3

n = 5 is a5+1/a5=a6/a5=8/5

Therefore, the required answer is 1, 2, 3/2, 5/3 and 8/5

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