# Tamilnadu Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

## Tamilnadu State Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.4

Question 1.
Find the next three terms of the following sequence.
(i) 8,24,72,…
(ii) 5,1,-3,…
(iii) $$\frac { 1 }{ 4 }$$,$$\frac { 2 }{ 9 }$$,$$\frac { 3 }{ 16 }$$………..
Solution:
(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,

∴ It is a geometric sequence
∴ The nth term of a G.P is tn = arn-1

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

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

Question 2.
Find the first four terms of the sequences whose nth terms are given by
(i) an = n3 – 2
(ii) an = (-1)n+1 n(n+1)
(iii) an = 2n2 – 6
Solution:
tn = an = n3 -2
(i) 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 first four terms are -1, 6, 25,62,…

(ii) an = (-1)n+1 n(n + 1)
a1 = (-1)1+1 (1) (1 +1)
= (-1)2 (1) (2) = 2
a2 = (-1)2+1 (2) (2 + 1)
= (-1)3 (2) (3)= -6
a3 = (-1)3+1 (3) (3 + 1)
= (-1)4 (3) (4) = 12
a4 = (-1)4+1 (4) (4 + 1)
= (-1)5 (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…

(iii) an = 2n2 – 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 first four terms are -4, 2, 12, 26, …

Question 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,…
Solution:
(i) 2,5,10,17
= 12 + 1, 22 + 1, 32 + 1, 42 + 1 …
∴ nth term is n2+1
(ii) 0, $$\frac { 1 }{ 2 }$$,$$\frac { 2 }{ 3 }$$,………….
= $$\frac { 1-1 }{ 1 }$$,$$\frac { 2-1 }{ 2 }$$,$$\frac { 3-1 }{ 3 }$$…..
⇒ $$\frac { n-1 }{ n }$$
∴ nth term is $$\frac { n-1 }{ n }$$
(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

Question 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
Solution:

Question 5.
Find a8 and a15 whose nth term is

Solution:

Question 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.
Solution:
a1 = 1, a2 = 1, an = 2an-1 + an-2
a3 = 2a(3-1) + a(3-2)
= 2a2 + a1
= 2 × 1 + 1 = 3
a4 = 20(4-1) + a(4-2)
= 2a3 + a2
= 2 × 3 + 1 = 7
a5 = 2a(5-1) + a(5-2)
= 2a4 + a3
= 2 × 7 + 3 = 17
a6 = 2a(6-1) + a(6-2)
= 2a5 + a74
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7,17,41……