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

In the given A.P., First term (a) = 3 and common difference (d) = 15 - 3 = 12. By formula, a n = a + (n - 1)d. Let nth term of A.P. be 132 more than 54th term. ∴ a n - a 54 = 132 ⇒ a + (n - 1)d - [a + (54 - 1)d] = 132 ⇒ 3 + 12(n - 1) - [3 + 53 × 12] = 132 ⇒ 3 + 12n - 12 - 3 - 636 = 132 ⇒ 12n - 648 = 132 ⇒ 12n = 132 + 648 ⇒ 12n = 780 ⇒ n = = 65. Hence, required term = 65th.