# 10th Maths Answers Chapter 2 Numbers and Sequences Ex 2.4

## 10th Maths Answers Chapter 2 Numbers and Sequences Ex 2.4

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1.Find the next three terms of the following sequence.

(i) 8, 24, 72, …….

(ii) 5, 1, -3, …….

(iii) \frac{1}{4}  \frac{2}{9} \frac{3}{16} ………..

(i) 8, 24, 72…

In an arithmetic sequence a = 8,

d = t1 – t1 = t3 – t2

= 24 – 8       72 – 24

= 16 ≠ 48

So, it is not an arithmetic sequence. In a geometric sequence,

r = \frac{t_{2}}{t_{1}} = \frac{t_{3}}{t_{2}} = \frac{t_{2}}{t_{1}} = \frac{t_{3}}{t_{2}}

⇒ \frac { 24 }{ 8 } = \frac { 72 }{ 24 }

⇒ 3 = 3

∴ It is a geometric sequence

∴ The nth term of a G.P is tn = arn-1

∴ t4 = 8 × 34-1

= 8 × 33

= 8 × 27

= 216

t5 = 8 × 35-1

= 8 × 34

= 8 × 81

= 648

t6 = 8 × 36-1

= 8 × 35

= 8 × 243

= 1944

The next 3 terms are 8, 24, 72, 216, 648, 1944.

(ii) 5, 1, -3, …

d = t2 – t1 = t3 – t2

⇒ 1 – 5 = -3-1

-4 = -4 ∴ It is an A.P.

tn = a+(n – 1)d

t4 = 5 + 3 × – 4

= 5 – 12

= -7

15 = a + 4d

= 5 + 4 × -4

= 5 – 16

= -11

t6 = a + 5d

= 5 + 5 × – 4

= 5 – 20

= – 15

∴ The next three terms are 5, 1, -3, -7, -11, -15.

(iii) \frac { 1 }{ 4 },\frac { 2 }{ 9 },\frac { 3 }{ 16 },………..

Here an = Numerators are natural numbers and denominators are squares of the next numbers

\frac { 1 }{ 4 },\frac { 2 }{ 9 },\frac { 3 }{ 16 } ,\frac { 4 }{ 25 },\frac { 5 }{ 36 },\frac { 6 }{ 49 }………….

2.Find the first four terms of the sequences whose nth terms are given by

(i) an = n3 -2

an = n3 – 2

a1 = 13 – 2 = 1 – 2 = -1

a2 = 23 – 2 = 8 – 2 = 6

a3 = 33 – 2 = 27 – 2 = 25

a4 = 43 – 2 = 64 – 2 = 62

The four terms are -1, 6, 25 and 62

(ii) an = (-1)n+1 n(n + 1)

an = (-1)n+1 n(n + 1)

a1 = (-1)2 (1) (2) = 1 × 1 × 2 = 2

a2 = (-1)3 (2) (3) = -1 × 2 × 3 = -6

a3 = (-1)4 (3) (4) = 1 × 3 × 4 = 12

a4 = (-1)5 (4) (5) = -1 × 4 × 5 = -20

The four terms are 2, -6, 12 and -20

(iii) an = 2n2 – 6

an = 2 n2 – 6

a1 = 2(1)2 – 6 = 2 – 6 = -4

a2 = 2(2)2 – 6 = 8 – 6 = 2

a3 = 2(3)2 – 6 = 18 – 6 = 12

a4 = 2(4)2 – 6 = 32 – 6 = 26

The four terms are -4, 2, 12, 26

3.Find the nth term of the following sequences

(i) 2, 5, 10, 17, ……….

(ii) 0, \frac { 1 }{ 2 }, \frac { 2 }{ 3 },…..

(iii) 3, 8, 13, 18, ……

(i) 2, 5, 10, 17

= 12 + 1, 22 + 1, 32 + 1, 42 + 1 ……….

∴ nth term is n2+1

(ii) 0, \frac { 1 }{ 2 },\frac { 2 }{ 3 },………….

= \frac { 1 - 1 }{ 1 },\frac { 2 - 1 }{ 2 },\frac { 3 - 1 }{ 3 }…..

⇒ \frac { n - 1 }{ 1 }

∴ nth term is \frac { n - 1 }{ 1 }

(iii) 3, 8, 13, 18

a = 3

d = 5

tn = a + (n – 1)d

= 3 + (n – 1)5

= 3 + 5n – 5

= 5n – 2

∴ nth term is 5n – 2

4.Find the indicated terms of the sequences whose nth terms are given by

(i) an = \frac { 5n }{ n + 2 } ; a6 and a13

(ii) an = -(n2 – 4); a4 and a11

5.Find a8 and a15 whose nth term is

6.If a1 = 1, a2 = 1 and an = 2an-1 + an-2 n > 3, n ∈ N. Then find the first six terms of the sequence.

a1 = a2 = 1

an = 2an-1 + an-2

a3 = 2a3-1 + a3-2 = 2a2 + a1

= 2(1) + 1 = 3

a4 = 2a4-1 + a4-2

= 2a3 + a2

= 2(3) + 1 = 6 + 1 = 7

a5 = 2 a5-1 + a5-2

= 2a4 + a3

= 2(7) + 3 = 17

a6 = 2a6-1 + a6-2

= 2a5 + a4

= 2(17) + 7

= 34 + 7 = 41

The sequence is 1, 1, 3, 7, 17,41, …