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.

Answer:

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 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = ss+2

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

a1 = 11+2 = 13

a2 = 22+2 = 24

a3 = 33+2 = 35

a4 = 44+2 = 46

a5 = 55+2 = 57

The starting five terms of the sequence are 12,23,34,45and56

 

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

Answer:

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 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = 2s+23

Putting s = 1, 2, 3, 4 and 5 respectively, in 2s+23

a1 = 2.1+23=43

a2 = 2.2+23=63

a3 = 2.3+23=83

a4 = 2.4+23=103

a5 = 2.5+23=123

The starting five terms of the sequence are 43,63,83,103and123

 

Q5: as = (1)s1+3s+1 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = (1)s1+3s+1

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s1+3s+1

a1 = (1)11+31+1 = 1 + 9 = 10

a2 = (1)21+32+1 = – 1 + 27 = 26

a3 = (1)31+33+1 = 1 + 81 = 82

a4 = (1)41+34+1 = – 1 + 243 = 242

a5 = (1)51+35+1 = 1 + 729 = 730

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

 

Q6: as = (1)s13s+1 is the sth term of a sequence. Find the starting five terms of the sequence.

Answer:

as = (1)s13s+1

Putting s = 1, 2, 3, 4 and 5 respectively, in (1)s13s+1

a1 = (1)1131+1=(1)032=19

a2 = (1)2132+1=(1)133=127

a3 = (1)3133+1=(1)234=181

a4 = (1)4134+1=(1)335=1243

a5 = (1)5135+1=(1)436=1729

The starting five terms of the sequence are 19,127,181,1243,and1729

 

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

Answer:

as = s + 3s

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

a17 = 17 + 317

a22 = 22 + 322

 

Q8: as = 9+ss2 is the sth term of a sequence. Obtain the 24th term in the required sequence.

Answer:

as = 9+ss2

Putting s = 24, in as = 9+ss2

a24 = 9+24242=33576

 

Q9: as = (1)s×6s is the sth term of a sequence. Obtain the 11th and 12th term in the required sequence.

Answer:

as = (1)s×6s

Putting s = 11 and 12, in as = (1)s×6s

a11 = (1)11×611=611

a12 (1)12×612=612

 

Q10. as = s+32+s is the sth term of a sequence. Obtain the 21st and 82nd term in the required sequence.

Answer:

as = s+32+s

Putting s = 21 and 82, in as = s+32+s

a21=21+32+21=2423 a82=82+32+82=8584

 

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

Answer:

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 for all s > 3

Answer:

a1 = 2, as = as112 for all s > 3

a2112=a112=212=12a3=a3112=a212=1212=14a4=a4112=a312=1412=58a5=a5112=a412=5812=1316

The starting five terms of the sequence are 2,12,14,58and1316

The following series is 2+12+(14)+(58)+(1316)+

 

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

Answer

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  for n = 2, 4.

Answer:

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 a4+1a4=a5a4=53