# Class 11 Maths Ncert Solutions Ex 9.1

## Class 11 Maths Ncert Solutions Chapter 9 Ex 9.1

Q1: as = s (s + 3) is the sth term of a sequence. Find the starting five terms of the sequence.

as = s (s + 3)

Putting s = 1, 2, 3, 4 and 5 respectively, in as = s (s + 3)

a1 = 1 (1 + 3) = 4

a2 = 2 (2 + 3) = 10

a3 = 3 (3 + 3) = 18

a4 = 4 (4 + 3) = 28

a5 = 5 (5 + 3) = 40

The starting five terms of the sequence are 4, 10, 18, 28 and 40

Q2: as = ss+2$\frac{s}{s + 2}$ is the sth term of a sequence. Find the starting five terms of the sequence.

as = ss+2$\frac{s}{s + 2}$

Putting s = 1, 2, 3, 4 and 5 respectively, in ss+2$\frac{s}{s + 2}$

a1 = 11+2$\frac{1}{1 + 2}$ = 13$\frac{1}{3}$

a2 = 22+2$\frac{2}{2 + 2}$ = 24$\frac{2}{4}$

a3 = 33+2$\frac{3}{3 + 2}$ = 35$\frac{3}{5}$

a4 = 44+2$\frac{4}{4 + 2}$ = 46$\frac{4}{6}$

a5 = 55+2$\frac{5}{5 + 2}$ = 57$\frac{5}{7}$

The starting five terms of the sequence are 12,23,34,45and56$\frac{1}{2},\; \frac{2}{3},\; \frac{3}{4},\; \frac{4}{5}\; and\; \frac{5}{6}$

Q3: as = 5s is the sth term of a sequence. Find the starting five terms of the sequence.

as = 5s

Putting s = 1, 2, 3, 4 and 5 respectively, in as = 5s

a1 = 51 = 5

a2 = 52 = 25

a3 = 53 = 125

a4 = 54 = 625

a5 = 55 = 3125

The starting five terms of the sequence are 5, 25, 125, 625 and 3125.

Q4: as = 2s+23$\frac{2s + 2}{3}$ is the sth term of a sequence. Find the starting five terms of the sequence.

as = 2s+23$\frac{2s + 2}{3}$

Putting s = 1, 2, 3, 4 and 5 respectively, in 2s+23$\frac{2s + 2}{3}$

a1 = 2.1+23=43$\frac{2.1 + 2}{3} = \frac{4}{3}$

a2 = 2.2+23=63$\frac{2.2 + 2}{3} = \frac{6}{3}$

a3 = 2.3+23=83$\frac{2.3 + 2}{3} = \frac{8}{3}$

a4 = 2.4+23=103$\frac{2.4 + 2}{3} = \frac{10}{3}$

a5 = 2.5+23=123$\frac{2.5 + 2}{3} = \frac{12}{3}$

The starting five terms of the sequence are 43,63,83,103and123$\frac{4}{3},\; \frac{6}{3},\; \frac{8}{3},\; \frac{10}{3}\; and\; \frac{12}{3}$

Q5: as = (1)s1+3s+1$(- 1)^{s – 1} + 3^{s + 1}$ is the sth term of a sequence. Find the starting five terms of the sequence.

as = (1)s1+3s+1$(- 1)^{s – 1} + 3^{s + 1}$

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s1+3s+1$(- 1)^{s – 1} + 3^{s + 1}$

a1 = (1)11+31+1$(- 1)^{1 – 1} + 3^{1 + 1}$ = 1 + 9 = 10

a2 = (1)21+32+1$(- 1)^{2 – 1} + 3^{2 + 1}$ = – 1 + 27 = 26

a3 = (1)31+33+1$(- 1)^{3 – 1} + 3^{3 + 1}$ = 1 + 81 = 82

a4 = (1)41+34+1$(- 1)^{4 – 1} + 3^{4 + 1}$ = – 1 + 243 = 242

a5 = (1)51+35+1$(- 1)^{5 – 1} + 3^{5 + 1}$ = 1 + 729 = 730

The starting five terms of the sequence are 10, 26, 82, 242 and 730.

Q6: as = (1)s13s+1$\frac{(- 1)^{s – 1}}{3^{s + 1}}$ is the sth term of a sequence. Find the starting five terms of the sequence.

as = (1)s13s+1$\frac{(- 1)^{s – 1}}{3^{s + 1}}$

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s13s+1$\frac{(- 1)^{s – 1}}{3^{s + 1}}$

a1 = (1)1131+1=(1)032=19$\frac{(- 1)^{1 – 1}}{3^{1 + 1}} = \frac{(- 1)^{0}}{3^{2}} = \frac{1}{9}$

a2 = (1)2132+1=(1)133=127$\frac{(- 1)^{2 – 1}}{3^{2 + 1}} = \frac{(- 1)^{1}}{3^{3}} = \frac{- 1}{27}$

a3 = (1)3133+1=(1)234=181$\frac{(- 1)^{3 – 1}}{3^{3 + 1}} = \frac{(- 1)^{2}}{3^{4}} = \frac{1}{81}$

a4 = (1)4134+1=(1)335=1243$\frac{(- 1)^{4 – 1}}{3^{4 + 1}} = \frac{(- 1)^{3}}{3^{5}} = \frac{- 1}{243}$

a5 = (1)5135+1=(1)436=1729$\frac{(- 1)^{5 – 1}}{3^{5 + 1}} = \frac{(- 1)^{4}}{3^{6}} = \frac{1}{729}$

The starting five terms of the sequence are 19,127,181,1243,and1729$\frac{1}{9},\; \frac{- 1}{27},\; \frac{1}{81},\; \frac{- 1}{243},\; and\; \frac{1}{729}$

Q7: as = s + 3s is the sth term of a sequence. Obtain the 17th and 22nd term in the required sequence.

as = s + 3s

Putting s = 17 and 22 respectively, in as = s + 3s

a17 = 17 + 317

a22 = 22 + 322

Q8: as = 9+ss2$\frac{9 + s}{s^{2}}$ is the sth term of a sequence. Obtain the 24th term in the required sequence.

as = 9+ss2$\frac{9 + s}{s^{2}}$

Putting s = 24, in as = 9+ss2$\frac{9 + s}{s^{2}}$

a24 = 9+24242=33576$\frac{9 + 24}{24^{2}} = \frac{33}{576}$

Q9: as = (1)s×6s$(- 1)^{s} \times 6^{s}$ is the sth term of a sequence. Obtain the 11th and 12th term in the required sequence.

as = (1)s×6s$(- 1)^{s} \times 6^{s}$

Putting s = 11 and 12, in as = (1)s×6s$(- 1)^{s} \times 6^{s}$

a11 = (1)11×611=611$(- 1)^{11} \times 6^{11} = – 6^{11}$

a12 (1)12×612=612$(- 1)^{12} \times 6^{12} = 6^{12}$

Q10. as = s+32+s$\frac{s + 3}{2 + s}$ is the sth term of a sequence. Obtain the 21st and 82nd term in the required sequence.

as = s+32+s$\frac{s + 3}{2 + s}$

Putting s = 21 and 82, in as = s+32+s$\frac{s + 3}{2 + s}$

a21=21+32+21=2423$a_{21} = \frac{21 + 3}{2 + 21} = \frac{24}{23}$ a82=82+32+82=8584$a_{82} = \frac{82 + 3}{2 + 82} = \frac{85}{84}$

Q11. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = 2 as – 1 for all s > 1

a1 = 2, as = 2 as – 1 for all s > 1

a2 = 2 a1 = 2 x 2 = 4

a3 = 2 a2 = 2 x 4 = 8

a4 = 2 a3 = 2 x 8 = 16

a5 = 2 a4 = 2 x 16 = 32

The starting five terms of the sequence are 2, 4, 8, 16 and 32

The following series is 2 + 4 + 8 + 16 + 32 + …….

Q12. Find out the starting five terms and also the following series of the required sequence given below:

a1 = 2, as = as112$\frac{a_{s – 1} – 1}{2}$ for all s > 3

a1 = 2, as = as112$\frac{a_{s – 1} – 1}{2}$ for all s > 3

a2112=a112=212=12a3=a3112=a212=1212=14a4=a4112=a312=1412=58a5=a5112=a412=5812=1316$\frac{a_{2 – 1} – 1}{2} = \frac{a_{1} – 1}{2} = \frac{2 – 1}{2} = \frac{1}{2} \\ a_{3} = \frac{a_{3 – 1} – 1}{2} = \frac{a_{2} – 1}{2} = \frac{\frac{1}{2} – 1}{2} = – \frac{1}{4} \\ a_{4} = \frac{a_{4 – 1} – 1}{2} = \frac{a_{3} – 1}{2} = \frac{- \frac{1}{4} – 1}{2} = – \frac{5}{8} \\ a_{5} = \frac{a_{5 – 1} – 1}{2} = \frac{a_{4} – 1}{2} = \frac{- \frac{5}{8} – 1}{2} = – \frac{13}{16} \\$

The starting five terms of the sequence are 2,12,14,58and1316$2,\; \frac{1}{2},\; – \frac{1}{4}, \; – \frac{5}{8}\; and\; – \frac{13}{16}$

The following series is 2+12+(14)+(58)+(1316)+$2\; + \frac{1}{2}\; + (- \frac{1}{4})\; + (- \frac{5}{8})\; +\; (- \frac{13}{16}) + ……$

Q13.  Find out the starting five terms and also the following series of the required sequence given below:

a1 = a2 = 4, as = as – 1 + 1 for all s > 4

a3 = a 3 – 1 + 1 = a2 +1 = 4 + 1 = 5

a4 = a 4 – 1 + 1 = a3 + 1 = 5 + 1 = 6

a5 = a 5 – 1 + 1 = a4 + 1 = 6 + 1 = 7

The starting five terms of the sequence are 4, 4, 5, 6 and 7

The following series is 4 + 4 + 5 + 6 + 7 + ….

Q14. The sequence of Fibonacci is described as 1 = a1 = a2 and as = as – 1 + as – 2, n > 1. Obtain as+1as$\frac{a_{s + 1}}{a_{s}}$  for n = 2, 4.

1 = a1 = a2 and as = as – 1 + as – 2, n > 1.

a3 = a2 + a1 = 1 + 1 = 2

a4 = a3 + a2 = 2 + 1 = 3

a5 = a4 + a3 = 3 + 2 = 5

a6 = a5 + a4 = 5 + 3 = 8

For s = 2 and 4 respectively,

a2+1a2=a3a2=21=2$\frac{a_{2 + 1}}{a_{2}} = \frac{a_{3}}{a_{2}} = \frac{2}{1} = 2$ a4+1a4=a5a4=53$\frac{a_{4 + 1}}{a_{4}} = \frac{a_{5}}{a_{4}} = \frac{5}{3}$