Find(i) 10th term of A.P. 2, 7, 12, ….(ii) 18th term of A.P. √2,

1.	Find(i) 10th term of A.P. 2, 7, 12, ….(ii) 18th term of A.P. √2, 3√2, 5√2, …..(iii) 24th term of A.P. 9, 13, 17, 21, ….
Answer» (i) Given A.P. 2, 7, 12, ….. First term a = 2 Common difference d = 7 – 2 = 5 n^th term a_n = a + (n – 1)d ∴ a₁₀ = 2 + (10 – 1) × 5 = 2 + 9 × 5 = 2 + 45 = 47 Hence, 10^th term of given series 47. (ii) Given A.P. √2, 3√2, 5√2, ….. First term a = √2 Common term d = 3√2 – √2 = √2(3 – 1) = 2√2 n^th term a_n = a + (n – 1)d ∴ a₁₈ = √2 + (18 – 1)2√2 = √2 + 17 × 2√2 = √2 + 34√2 = 35√2 Hence a₁₈ = 35√2 (iii) Given A.P. 9, 13, 17, 21, ….. First term a = 9 common difference d = 13 – 9 = 4 n^th term a_n = a + (n – 1)d ∴ a₂₄ = 9 + (24 – 1) × 4 = 9 + 23 × 4 = 9 + 92 = 101 Hence a₂₄ = 101

Find(i) 10th term of A.P. 2, 7, 12, ….(ii) 18th term of A.P. √2, 3√2, 5√2, …..(iii) 24th term of A.P. 9, 13, 17, 21, ….

Answer»

(i) Given A.P. 2, 7, 12, …..

First term a = 2

Common difference d = 7 – 2 = 5

n^th term a_n = a + (n – 1)d

∴ a₁₀ = 2 + (10 – 1) × 5

= 2 + 9 × 5

= 2 + 45

= 47

Hence, 10^th term of given series 47.

(ii) Given A.P. √2, 3√2, 5√2, …..

First term a = √2

Common term d = 3√2 – √2

= √2(3 – 1) = 2√2

n^th term a_n = a + (n – 1)d

∴ a₁₈ = √2 + (18 – 1)2√2

= √2 + 17 × 2√2

= √2 + 34√2

= 35√2

Hence a₁₈ = 35√2

(iii) Given A.P. 9, 13, 17, 21, …..

First term a = 9

common difference

d = 13 – 9 = 4

n^th term a_n = a + (n – 1)d

∴ a₂₄ = 9 + (24 – 1) × 4

= 9 + 23 × 4

= 9 + 92

= 101

Hence a₂₄ = 101