Question 11:

Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54th term?

Answer:

Given A.P. is 3, 15, 27, 39, …

a = 3

d = a₂ − a₁ = 15 − 3 = 12

a₅₄ = a + (54 − 1) d

= 3 + (53) (12)

= 3 + 636 = 639

132 + 639 = 771

We have to find the term of this A.P. which is 771.

Let n^th term be 771

a_n = a + (n − 1) d

771 = 3 + (n − 1) 12

768 = (n − 1) 12

(n − 1) = 64

n = 65

Therefore, 65^th term was 132 more than 54^th term.

Alternatively,

Let n^th term be 132 more than 54th term.

n = 54 + (132/12)

= 54 + 11 = 65^th term

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