Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

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

Solutions to questions 1 – 6:

1. a_{n} = n(n + 2)

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

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

4. a_{n} = \(\frac{2n-3}{6}\)

5. a_{n} = (- 1)^{n-1}5^{n+1}

6. a_{n} = \(\frac{n\left(n^{2}+5\right)}{4}\)

Solution to questions 1 – 6:

1. a_{n} = n(n + 2)

Putting n = 1, 2, 3, 4 and 5, we get

a_{1} = 1(1 + 2) = 1.3 = 3, a_{2} = 2.(2 + 2) = 2.4 = 8,

a_{3} = 3(3 + 2) = 3.5= 15, a_{4} = 4.(4 + 2) = 4.6 = 24 and

a_{5} = 5(5 + 2) = 5.7 = 35.

∴ The first of five terms of the sequence, whose nth term is

a_{n} = n(n + 2) are 3, 8, 15, 24 and 35.

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

Putting n = 1, 2, 3, 4 and 5, we get

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

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

and a_{5} = \(\frac{5}{5+1}\) = \(\frac{5}{6}\).

∴ The first five terms of the sequence whose

nth term is \(\frac{n}{n+1}\) are \(\frac{1}{2}\), \(\frac{2}{3}\), \(\frac{3}{4}\), \(\frac{4}{5}\) and \(\frac{5}{6}\).

3. a_{n} = 2^{n}. Putting n = 1, 2, 3, 4 and 5, we get

a_{1} = 2^{1} = 2, a_{2} = 4, a_{3} = 2^{3} = 8,

a_{4} = 2^{4} = 16 and a_{5} = 2^{5} = 32.

∴ Required first five terms of the sequence are 2, 4, 8, 16 and 32.

4. Here, a_{n} = \(\frac{2n-3}{6}\).

Putting n = 1, 2, 3, 4 and 5, we get

a_{1} = \(\frac{2×1-3}{6}\) = \(\frac{2-3}{6}\) = \(\frac{-1}{6}\);

a_{2} = \(\frac{2×2-3}{6}\) = \(\frac{4-3}{6}\) = \(\frac{1}{6}\);

a_{3} = \(\frac{2×3-3}{6}\) = \(\frac{6-3}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);

a_{4} = \(\frac{2×4-3}{6}\) = \(\frac{8-3}{6}\) = \(\frac{5}{6}\)

and a_{5} = \(\frac{2×5-3}{6}\) = \(\frac{10-3}{6}\) = \(\frac{7}{6}\).

5. a_{n} = (- 1)^{n-1}5^{n+1}

a_{1} = (- 1)^{0}.5^{i+1} = 5^{2} = 25,

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

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

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

and a_{5} = (- 1)^{5-1}.5^{5+1} = 5^{6} = 15625.

The first five terms of the given sequence are

25, – 125, 625, – 3125 and 15625.

6. a_{n} = \(\frac{n\left(n^{2}+5\right)}{4}\)

Putting n = 1, 2, 3, 4 and 5, we get

∴ The first five terms are \(\frac{3}{2}\), \(\frac{9}{2}\) \(\frac{21}{2}\), 21 and \(\frac{75}{2}\).

Find the indicated terms in each of the following sequences in questions 7 to 10 whose nth terms are

7. a_{n} = 4n – 3; a_{17}, a_{24}

8. a_{n} = \(\frac{n^{2}}{2^{n}}\); a_{7}

9. a_{n} = (- 1)^{n-1}n^{3}, a_{9}

10. \(\frac{n(n-2)}{n+3}\); a_{20}

Solutions to questions 7 to 10:

7. a_{n} = 4n – 3, Putting n = 17 and 24, we get

a_{17} = 4.17 – 3 = 68 – 3 = 65

and a_{24} = 4.24 – 3 = 96 – 3 = 93.

8. a_{n} = \(\frac{n^{2}}{2^{n}}\); a_{7} Putting n = 7, we get

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

9. a_{n} = (- 1)^{n-1}n^{3}, Putting n = 9, we get

a_{n} = (- 1)^{9-1}9^{3} = 729.

10. a_{n} = \(\frac{n(n-2)}{n+3}\). Putting n = 20, we get

a_{20} = \(\frac{20(20-2)}{20+3}\) = \(\frac{20.18}{23}\) = \(\frac{360}{23}\).

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

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

12. a_{1} = – 1, a_{n} = \(\frac{a_{n-1}}{n}\), n ≥ 2

13. a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1, n ≥ 2

Solution to questions 11 to 13:

11. a_{1} = 3, a_{n} = 3a_{n-1} + 2.

Putting n = 2, 3, 4 and 5, we get

a_{2} = 3a_{1} + 2 = 3.3 + 2 = 9 + 2 = 11,

a_{3} = 3a_{2} + 2 = 3.11 + 2 = 33 + 2 = 35,

a_{4} = 3a_{3} + 2 = 3.35 + 2 = 105 + 2 = 107.

and a_{5} = 3a_{4} + 2 = 3.107 + 2 = 321 + 2 = 323.

The first five terms of the sequence are 3, 11, 35, 107 and 323.

∴ The corresponding series is 3 + 11 + 35 + 107 + 323 + ……………..

12. Here, a_{1} = – 1, a_{n} = \(\frac{a_{n-1}}{n}\), n ≥ 2.

Putting n = 2, 3, 4, 5 and 6, we get

∴ The first five terms of the sequence are – 1, – \(\frac{1}{2}\), – \(\frac{1}{6}\), – \(\frac{1}{24}\) and \(\frac{- 1}{120}\).

The corresponding series is – 1, – \(\frac{1}{2}\) – \(\frac{1}{6}\) – \(\frac{1}{24}\) – \(\frac{1}{120}\) – …………………

13. Here, a_{1} = a_{2} = 2, a_{n} = a_{n-1} – 1 (n > 2)

Putting n = 3, 4, 5, 6 and 7, we get

a_{3} = a_{3-1} – 1 = a_{2} – 1 = 2 – 1 = 1;

a_{4} = a_{4-1} – 1 = a_{3} – 1 = 1 – 1 = 0;

a_{5} = a_{5-1} – 1 = a_{4} – 1 = 0 – 1 = – 1;

The first five terms of the sequences are 2, 2, 1, 0 and – 1.

The corresponding series is 2 + 2 + 1 + 0 – 1 + …………………

14. The fibonacci sequence is defined by 1 = a_{1} = a_{2} and a_{n} = a_{n-1} + a_{n-2}, n > 2.

Find \(\frac{a_{n+1}}{a_{n}}\) for n = 1, 2, 3, 4 and 5.

Solution:

Here, a_{1} = 1 = a_{2} and a_{n} = a_{n-1} + a_{n-2} (n > 2) …………….. (A)

The values of \(\frac{a_{n+1}}{a_{n}}\) for n = 1, 2, 3, 4 and 5 are respectively 1, 2, \(\frac{3}{2}\), \(\frac{5}{3}\) and \(\frac{8}{5}\).