Class 11 NCERT Solutions Chapter 9 Sequences And Series Exercise 9.1 (original) (raw)

Last Updated : 11 Sep, 2024

Chapter 9 of the NCERT Class 11 Mathematics textbook covers Sequences and Series which are fundamental concepts in algebra and mathematical analysis. Exercise 9.1 specifically deals with the problems that help students understand and solve various types of sequences and series. Mastery of these problems is crucial for developing a strong foundation in higher mathematics and for tackling more complex topics in future studies.

What are Sequences and Series?

The Sequences are ordered lists of numbers where each term is determined by the specific rule or pattern. For example, in the sequence 2,4,6,8,… each term increases by 2. Series refers to the sum of the terms in a sequence. For instance, the series corresponding to the sequence above would be 2+4+6+8+⋯. There are various types of sequences and series including arithmetic sequences and geometric sequences.

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

**Solution:

Given, **a n **= 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, a n = 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, **a n =2 n

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, a n =(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, **a n = (-1) n-1 5 n+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, a n **= 4n - 3

Putting value of n as 17. We get,

a17 = 4(17) -3 = 68 - 3=65

**Therefore, the 17 th 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 7 th 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 9 th 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, **a n **= (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 20 th 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

Conclusion

Understanding sequences and series is essential for the solving a wide range of mathematical problems and for the applications in the various fields including finance, computer science and engineering. Exercise 9.1 provides the practice in identifying and solving problems related to these concepts helping students gain confidence and proficiency in working with the sequences and series.