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 - a₅₄ = 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 = $\dfrac{780}{12}$ = 65.

Hence, required term = 65th.