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

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
Then, 132 more than its 54th term is 132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let nth term be 771.
⇒ aₙ = a + (n - 1)d
⇒ 771 = 3 + (n - 1)12
⇒ 768 = (n - 1)12
⇒ (n - 1) = 64
∴ n = 65
Therefore, 65th term was 132 more than 54th term.