First term = 3 and common difference = 15 – 3 = 12
Let the nth term of AP: 3, 15, 27, 39 … will be 132 more than its 54th term.
Therefore, a_n = a₅₄ + 132
⇒ a + (n -1)d = a + 53d + 132
⇒ (n -1)(12) = 53 × 12 + 132
⇒ (n -1)(12) = 768
⇒ n -1 = 768/12 = 64
⇒ n = 65
Hence, 65th term of the AP: 3, 15, 27, 39 . . . will be 132 more than its 54th trem.