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

a_{1} = 1 (1 + 2) = 3,

a_{2} = 2 (2 + 2) = 8,

a_{3} = 3 (3 + 2) = 15,

a_{4} = 4 (4 + 2) = 24,

a_{5} = 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} = 2^{n}.

Answer.

a_{n} = 2^{n}

Substituting n = 1, 2, 3, 4, 5, we obtain

a_{1} = 2^{1} = 2

a_{2} = 2^{2} = 4

a_{3} = 2^{3} = 8

a_{4} = 2^{4} = 16

a_{5} = 2^{5} = 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

a_{1} = (- 1)^{1 – 1} 5^{1 + 1} = 5^{2} = 25,

a_{2} = (- 1)^{2 – 1} 5^{2 + 1} = – 5^{3} = -125,

a_{3} = (- 1)^{3 – 1} 5^{3 + 1} = 5^{4} = 625,

a_{4} = (- 1)^{4 – 1} 5^{4 + 1} = 5^{5} = – 3125,

a_{5} = (- 1)^{5 – 1} 5^{5 + 1} = 5^{6} = 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; a_{17}, a_{24}.

Answer.

We have a_{n} = 4n -3

On putting n =17, we get

a_{17} = 4 × 17 – 3

= 68 – 3 = 65

On putting n = 24, we get

a_{24} = 4 × 24 – 3

= 96 – 3 = 93.

Question 8.

Find the indicate term in the following sequence whose nth term is a_{n} = \(\frac{n^{2}}{2^{n}}\); a_{7}.

Answer.

Substituting n = 7, we obtain

a_{7} = \(\frac{7^{2}}{2^{7}}=\frac{49}{128}\).

Question 9.

Find the indicated term in the following sequence whose nt1 term is a_{n} = (- 1)^{n – 1} n^{3}; a_{9}.

Answer.

Substituting n = 9, we obtain

a_{9} = (- 1)^{9 – 1} (9)^{3} = 729.

Question 10.

Find the indicated term in the following sequence whose nth term is a_{n} = \(\frac{n(n-2)}{n+3}\); a_{20}.

Answer.

Substituting n = 20, we obtain

a_{20} = \(\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\).

Question 11.

Write the first five terms of the following sequence and obtain the corresponding series:

a_{1} = 3, a_{n} = 3a_{n – 1} + 2 for all n > 1.

Answer.

a_{1} = 3, a_{n} = 3 a_{n – 1} + 2 for all n > 1

=> a_{2} = 3a_{2 – 1} + 2

= 3a_{1} + 2

= 3(3) + 2 = 11

a_{3} = 3a_{3 – 1} + 2

= 3a_{2} + 2

= 3(11) + 2 = 35

a_{4} = 3a_{4 – 1} + 2

= 3a_{3} + 2

= 3(35) + 2 = 107

a_{5} = 3a_{5 – 1} + 2

= 3a_{4} + 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:

a_{1} = – 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:

a_{1} = a_{2} = 2, a_{n} = a_{n – 1} – 1, n > 2.

Answer.

a_{1} = a_{2} = 2,

a_{n} = a_{n – 1}, n > 2

=> a_{3} = a_{2} – 1 = 2 – 1 = 1,

a_{4} = a_{3} – 1 = 1 – 1 = 0

a_{5} = a_{4} – 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 = a_{1} = a_{2} 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 = a_{1} = a_{2}

a_{n} = a_{n – 1} + a_{n – 2}, n > 2

a_{3} = a_{2} + a_{1}

= 1 + 1 = 2,

a_{4} = a_{3} + a_{2}

= 2 + 1 = 3,

a_{5} = a_{4} + a_{3}

= 3 + 2 = 5,

a_{6} = a_{5} + a_{3}

= 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,

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