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 n^th term of a G.P is tn = ar^n-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.
t_n 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 a_n = 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) a_n = n³ – 2
(ii) a_n = (-1)ⁿ⁺¹ n(n+1)
(iii) a_n = 2n² – 6
Solution:
t_n = a_n = 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) a_n = (-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) a_n = 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 n^th 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 …
∴ n^th 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
t_n = a + (n – 1)d
= 3 + (n – 1)5
= 3 + 5n – 5
= 5n – 2
∴ n^th term is 5n – 2

Question 4.
Find the indicated terms of the sequences whose n^th terms are given by
(i) a_n = \(\frac { 5n }{ n+2 } \) ; a₆ and a₁₃
(ii) a_n = -(n² – 4); a₄ and a₁₁
Solution:

Question 5.
Find a₈ and a₁₅ whose n^th term is

Solution:

Question 6.
If a₁ = 1, a₂ = 1 and an = 2a_n-1 + a_n-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a₁ = 1, a₂ = 1, a_n = 2a_n-1 + a_n-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……

Samacheer Kalvi 10th Maths Book Solutions