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## Tamilnadu Samacheer Kalvi 10th 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) $$\frac { 1 }{ 4 }$$, $$\frac { 2 }{ 9 }$$, $$\frac { 3 }{ 16 }$$
(i) 216, 648, 1944 (This sequence is multiple of 3)
Next three terms are 216, 648, 1944
(ii) Next three terms are -7, -11, -15
(iii) Next three terms are $$\frac { 4 }{ 25 }$$,$$\frac { 5 }{ 36 }$$ and $$\frac { 6 }{ 49 }$$
[using $$\frac{n}{(n+1)^{2}}$$]

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, ……
(12 + 1);(22 + 1),(32 + 1),(42 + 1)….
nth term is n2 + 1
an = n2 + 1

(ii) 0,$$\frac { 1 }{ 2 }$$,$$\frac { 2 }{ 3 }$$ ……
($$\frac { 1-1 }{ 1 }$$), ($$\frac { 2-1 }{ 2 }$$), ($$\frac { 3-1 }{ 3 }$$) …..
nth term is $$\frac { n-1 }{ n }$$
an = $$\frac { n-1 }{ n }$$

(iii) 3,8,13,18,…….
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The nth term is 5n – 2
an = 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
an = $$\frac { 5n }{ n+2 }$$
a6 = $$\frac { 5(6) }{ 6+2 }$$ = $$\frac { 30 }{ 8 }$$ = $$\frac { 15 }{ 4 }$$
a13 = $$\frac { 5(13) }{ 13+2 }$$ = $$\frac{5 \times 13}{15}$$ = $$\frac { 13 }{ 3 }$$
a6 = $$\frac { 15 }{ 4 }$$, a13 = $$\frac { 13 }{ 3 }$$

(ii) an = – (n2 – 4); a4 and a11
an = -(n2 – 4)
a4 = -(42 – 4)
= – (16 – 4)
= -12
a11 = -(112 – 4)
= – (121 – 4)
= – 117
a4 = -12 and a11 = -117

Question 5.
Find a8 and a15 whose nth term is an

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 = 2a(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 + a4
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..