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

Given: 3, 15, 27, 39, …

First we need to calculate 54^th term.

We know that

a_n = a + (n – 1)d

Here, a = 3, d = 15 – 3 = 12 and n = 54

So, a₅₄ = 3 + (54 – 1)12

⇒ a₅₄ = 3 + 53 × 12

⇒ a₅₄ = 3 + 636

⇒ a₅₄ = 639

Now, the term is 132 more than a₅₄ is 132 + 639 = 771

Now,

a + (n – 1)d = 771

⇒ 3 + (n – 1)12 = 771

⇒ 3 + 12n – 12 = 771

⇒ 12n = 771 + 12 – 3

⇒ 12n = 780

⇒ n = 65

Hence, the 65^th term is 132 more than the 54^th term.