RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 11 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Maths Important Questions for exam preparation. Students can also go through RBSE Class 11 Maths Notes to understand and remember the concepts easily.

RBSE Class 11 Maths Solutions Chapter 9 Sequences and Series Ex 9.1

Write the first five terms of each of the sequences in exercises 1 to 6 whose nth terms are:

Question 1.
an = n(n + 2)
Answer:
Putting n = 1, 2, 3, 4, 5
a1 = 1 × (1 + 2) = 3
a2 = 2 × (2 + 2) = 8
a3 = 3 × (3 + 2) = 15
a4 = 4 × (4 + 2) = 24
a5 = 5 × (5 + 2) = 35
Thus, first five terms of the sequence are 3, 8, 15, 24, 35.

RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Question 2.
an = \(\frac{n}{n+1}\)
Answer:
Putting n = 1, 2, 3, 4, 5
RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 1
Thus, first five terms of the sequence are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}\)

Question 3.
an = 2n
Answer:
Putting n = 1, 2, 3, 4, 5
a1 = 21 = 2
a2 = 22 = 4
a3 = 23 = 8
a4 = 24 = 16
a5 = 25 = 32
Thus, first five terms of the sequence are 2, 4, 8, 16, 32

Question 4.
an = \(\frac{2 n-3}{6}\)
Answer:
Putting n = 1, 2, 3, 4, 5
RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 2
Thus, first five terms of the sequence are \(\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}\).

RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Question 5.
an = (- 1)n - 1 5n + 1
Answer:
Putting n = 1, 2, 3, 4, 5
a1 = (- 1)1 - 1 51 + 1
= (- 1)0 52 = 25
a2 = (- 1)2 52 + 1
= (- 1)1 53 = - 125
a3 = (- 1)3 - 1 53 + 1
= (- 1)2 54 = 625
a4 = (- 1)4 - 1 54 + 1
= (- 1)3 55 = - 3125
a5 = (- 1)5 - 1 55 + 1
= (- 1)4 56 = 15625
Thus, first five terms of the sequence are 25, - 125, 625, - 3125, 15625

Question 6.
an = n\(\frac{n^2+5}{4}\)
Answer:
Putting n = 1, 2, 3, 4, 5
RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 3
Thus, five terms of the sequence are \(\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21, \frac{75}{2}\)

RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Find the indicated terms in each of the sequences in exercise 7 to 10 whose nth term are:

Question 7.
an = 4n - 3;a17, a24
Answer:
nth term of given the sequence an = 4n - 3
Putting n = 17 and n = 24 in an = 4n - 3
a17 = 4 × 17 - 3
= 68 - 3
= 65
and a24 = 4 × 24 - 3
= 96 - 3
= 93
Thus, a17 = 65 and a24 = 93

Question 8.
an = \(\frac{n^2}{2^n}\); a7
Answer:
nth term of the given sequence an = \(\frac{n^2}{2^n}\)
Putting n = 7, we get
a7 = \(\frac{7^2}{2^7}\) = \(\frac{49}{128}\)
Thus, the required term a7 = \(\frac{49}{128}\)

Question 9.
a = (- 1)n - 1 n3; a9
Answer:
nth term of the given sequence
an = (- 1)n - 1 n3
Putting n = 9, we get
a = (- 1)9 - 1 93
= (- 1)8 93 = 729
Thus, required term a9 = 729

Question 10.
an = \(\frac{n(n-2)}{n+3}\); a20
Answer:
nth term of the given sequence
an = \(\frac{n(n-2)}{n+3}\)
Putting n = 20, we get
a20 = \(\frac{20(20-2)}{20+3}=\frac{20 \times 18}{23}=\frac{360}{23}\)
Thus, the required term a20 = \(\frac{360}{23}\)

RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Write the first five terms of each of the sequences in exercise 11 to 13 and obtain the corresponding series.

Question 11.
a1 = 3, an = 3an - 1 + 2, for all n > 1.
Answer:
Putting n = 2, 3, 4, 5 in an = 3an - 1 + 2
a2 = 3a2 - 1 + 2
= 3a1 + 2
= 3 × 3+ 2 (∵ a1 = 3)
= 11

a3 = 3 × a3 - 1 + 2
= 3a2 + 2
= 3 × 11 + 2 (∵ a2 = 11)
= 35

a4 = 3a4 - 1 + 2
= 3 × a3 + 2
= 3 × 35 + 2 (∵ a3 = 35)
105 + 2 = 107

a5 = 3 × a5 - 1 + 2
= 3 × a4 + 2
= 3 × 107 + 2 (∵ a4 = 107)
= 321 + 2 = 323
Thus, five terms of sequence are 3, 11, 35, 107, 323 and corresponding series is 3 + 11 + 35 + 107 + 323....

Question 12.
a1 = - 1, an = \(\frac{a_{n-1}}{n}\), where n ≥ 2
Answer:
RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 4

RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Question 13.
a1 = a2 = 2, an = an - 1 - 1, n > 2
Answer:
Given a1 = a2 = 2, an = an - 1 - 1
Here, first term a1 = second term a2 = 2
Putting n = 3, 4, 5 in an = an - 1
a3 = a3 - 1 - 1
= a2 - 1
∴ a3 = 2 - 1 = 1 (∵ a2 = 2)
a4 = a4 - 1
= 1 - 1 (∵ a3 = 1)
= 0
a5 = a5 - 1 - 1
= a4 - 1 .
= 0 - 1 [∵ a4 = 0]
= - 1
Thus, five terms of sequence are 2, 2, 1, 0, - 1 and corresponding series is 2 + 2 + 1 + 0 + (- 1) + ..................

Question 14.
The Fibonacci sequence is defined by:
1 = a1 = a2 and an = an - 1 + an - 2, n > 2 find \(\frac{a_{n+1}}{a_n}\) for n = 1, 2, 3, 4, 5, ..............
Answer:
Given: 1 = a1 = a2 and an = an - 1 + an - 2
(Fibonacci sequence)
For n = 3, 4, 5, 6
nth term an = an - 1 + an - 2
Putting n = 3, we get
a3 = a3 - 1 + a3 - 2
= a2 + a1
= 1 + 1 (∵ a1 = a2 = 1)
= 2
Putting n = 4, we get
a4 = a4 - 1 + a4 - 2
= a3 + a2
= 2 + 1 (∵ a3 = 2, a2 = 1)
= 3
Putting n = 5, we get
a5 = a5 - 1 + a5 - 2
= a4 + a3
= 3 + 2 (∵ a4 = 3, a3 = 2)
= 5
And Putting n = 6, we get
a6 = a6 - 1 + a6 - 2
= a5 + a4
= 5 + 3 (∵ a5 = 5, a4 = 3)
= 8
Now,
RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 1

Bhagya
Last Updated on Nov. 13, 2023, 5:57 p.m.
Published Dec. 16, 2022