Solution:

Given to us that A.P. is 3, 15, 27, 39, …

The first term, a = 3

The common difference, d = a₂ − a₁ = 15 − 3 = 12

We all know that,

a_n = a+(n−1)d

Therefore,

a₅₄ = a+(54−1)d

⇒3+(53)(12)

⇒3+636 = 639

a₅₄ = 639

We need to find the term of the above A.P. which is 132 more than a_54, i.e.771.

Let n^th term be 771.

a_n = a+(n−1)d

771 = 3+(n −1)12

768 = (n−1)12

(n −1) = 64

n = 65

As a result, 65^th term was 132 more than 54^th term.

Another method is;
Let the n^th term should be 132 more than the 54^th term.

n = 54 + 132/2 = 54 + 11 = 65^th term