PSEB 11th Class Maths Solutions Chapter 9 Sequences and Series Ex 9.1

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Ex 9.1 Textbook Exercise Questions and Answers.

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

Question 1.
Write the first five terms of the sequence whose nth term is a_n = n (n + 2).
Answer.
a_n = n(n +2)
Substituting n = 1, 2, 3, 4 and 5, we obtain
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
Therefore, the required terms are 3, 8, 15, 24 and 35.

Question 2.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{n}{n+1}\).
Answer.
a_n = \(\frac{n}{n+1}\)
Sustituting n = 1, 2, 3, 4, 5, we otain
a_n = \(\frac{1}{1+1}=\frac{1}{2}\)

a_n = \(\frac{2}{2+1}=\frac{2}{3}\)

a_n = \(\frac{3}{3+1}=\frac{3}{4}\)

a_n = \(\frac{4}{4+1}=\frac{4}{5}\)

a_n = \(\frac{5}{5+1}=\frac{5}{6}\)

Therefore, the required terms are \(\frac{1}{2}\), \(\frac{2}{3}\) , \(\frac{3}{4}\) , \(\frac{4}{5}\) and \(\frac{5}{6}\) .

Question 3.
Write the first five terms of the sequence whose nth term is a_n = 2ⁿ.
Answer.
a_n = 2ⁿ
Substituting n = 1, 2, 3, 4, 5, we obtain
a₁ = 2¹ = 2
a₂ = 2² = 4
a₃ = 2³ = 8
a₄ = 2⁴ = 16
a₅ = 2⁵ = 32
Therefore, the required terms are 2, 4, 8, 16 and 32.

Question 4.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{2 n-3}{6}\).
Answer.
Substituting n = 1,2, 3, 4, 5, we obtain

Question 5.
WrIte the first five terms of the sequence whose nth term is
a_n = (- 1)^{n – 1} 5^{n + 1}.
Ans.
Substituting n = 1,2, 3, 4, 5, we obtain
a₁ = (- 1)^{1 – 1} 5^{1 + 1} = 5² = 25,
a₂ = (- 1)^{2 – 1} 5^{2 + 1} = – 5³ = -125,
a₃ = (- 1)^{3 – 1} 5^{3 + 1} = 5⁴ = 625,
a₄ = (- 1)^{4 – 1} 5^{4 + 1} = 5⁵ = – 3125,
a₅ = (- 1)^{5 – 1} 5^{5 + 1} = 5⁶ = 15625
Therefore, the required terms are 25, – 125, 625, – 3125, and 15625.

Question 6.
Write the first five terms of the sequence whose nth term is a_n = \(n \frac{n^{2}+5}{4}\).
Answer.
Substituting n = 1, 2, 3, 4, 5, we obtain

Question 7.
Find the indicated terms in the following sequence whose nth term is a_n =4n – 3; a₁₇, a₂₄.
Answer.
We have a_n = 4n -3
On putting n =17, we get
a₁₇ = 4 × 17 – 3
= 68 – 3 = 65
On putting n = 24, we get
a₂₄ = 4 × 24 – 3
= 96 – 3 = 93.

Question 8.
Find the indicate term in the following sequence whose nth term is a_n = \(\frac{n^{2}}{2^{n}}\); a₇.
Answer.
Substituting n = 7, we obtain
a₇ = \(\frac{7^{2}}{2^{7}}=\frac{49}{128}\).

Question 9.
Find the indicated term in the following sequence whose nt1 term is a_n = (- 1)^{n – 1} n³; a₉.
Answer.
Substituting n = 9, we obtain
a₉ = (- 1)^{9 – 1} (9)³ = 729.

Question 10.
Find the indicated term in the following sequence whose nth term is a_n = \(\frac{n(n-2)}{n+3}\); a₂₀.
Answer.
Substituting n = 20, we obtain
a₂₀ = \(\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\).

Question 11.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = 3, a_n = 3a_{n – 1} + 2 for all n > 1.
Answer.
a₁ = 3, a_n = 3 a_{n – 1} + 2 for all n > 1
=> a₂ = 3a_{2 – 1} + 2
= 3a₁ + 2
= 3(3) + 2 = 11

a₃ = 3a_{3 – 1} + 2
= 3a₂ + 2
= 3(11) + 2 = 35

a₄ = 3a_{4 – 1} + 2
= 3a₃ + 2
= 3(35) + 2 = 107

a₅ = 3a_{5 – 1} + 2
= 3a₄ + 2
= 3(107) + 2 = 323.
Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Question 12.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = – 1, a_n = \(\frac{\boldsymbol{a}_{n-1}}{n}\), n > 2
Answer.

Question 13.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = a₂ = 2, a_n = a_{n – 1} – 1, n > 2.
Answer.
a₁ = a₂ = 2,
a_n = a_{n – 1}, n > 2
=> a₃ = a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0
a₅ = a₄ – 1 = 0 – 1 = – 1.
Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1.
The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….

Question 14.
The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2} n > 2. Find \(\frac{\boldsymbol{a}_{n+1}}{\boldsymbol{a}_{n}}\), for n = 1, 2, 3, 4, 5.
Answer.
1 = a₁ = a₂
a_n = a_{n – 1} + a_{n – 2}, n > 2
a₃ = a₂ + a₁
= 1 + 1 = 2,
a₄ = a₃ + a₂
= 2 + 1 = 3,
a₅ = a₄ + a₃
= 3 + 2 = 5,
a₆ = a₅ + a₃
= 5 + 3 = 8

For n = 1,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}\) = 1

For n = 2,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}\) = 2

For n = 3,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}\)

For n = 4,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}\)

For n = 5,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}\)