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

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 n^th term is a_n=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 n^th term is a_n = 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 n^th term is a_n = 2ⁿ.

Solution:

Given, a_n=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 n^th term is a_n = (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 n^th term is a_n = (-1)^n-15ⁿ⁺¹.

Solution:

Given, a_n = (-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 n^th term is given as a_n = n(n²+5)/4.

Solution:

Given, a_n = 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 17^th term of the sequence whose n^th term is given as a_n=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 7^th term of the sequence whose n^th term is given as a_n = 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 9^th term of the sequence whose n^th term is given as a_n = (-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 20^th term of the sequence whose n^th term is gives as a_n = (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, a_n = 3a_n-1+ 2 for all n>1.

Solution:

a₁ = 3

a₂ = 3a_2-1+2=3a₁+2=3*3+2=9+2=11

a₃ = 3a_3-1+2=3a₂+2=3*11+2=33+2=35

a₄ = 3a_4-1+2=3a₃+2=3*35+2=105+2=107

a₅ = 3a_5-1+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, a_n = a_n-1/n for all n>1.

Solution:

a₁=-1

a₂=a_2-1/2=a₁/2=-1/2

a₃=a_3-1/3=a₂/3=(-1/2)/3=-1/6

a₄=a_4-1/4=a₃/4=(-1/6)/4=-1/24

a₅=a_5-1/5=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, a_n = a_n-1-1 for all n > 2.

Solution:

a₁ = 2

a₂ = 2

a₃ = a_3-1-1=a₂-1=2-1=1

a₄ = a_4-1-1=a₃-1=1-1=0

a₅ = a_5-1-1=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 a_n = a_n-1 + a_n-2, n > 2. Find a_n+1/a_n 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 a_n+1/a_n 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.