**Solution:**

Given AP is 3, 15, 27, 39, …

First term, a = 3

Common difference, d = a_{2} − a_{1} = 15 − 3 = 12

We know that,

a_{n} = a + (n − 1)d

Therefore,

a_{54} = a + (54 − 1) d

⇒ 3 + (53)(12)

⇒ 3 + 636 = 639

a_{54 }= 639 + 132 = 771

We have to find the term of this AP which is 132 more than a_{54, }i.e., 771.

Let 771 be the nth term of the AP.

a_{n} = a + (n − 1) d

771 = 3 + (n − 1)12

768 = (n −1)12

(n −1) = 64

n = 65

Therefore, the 65^{th} term was 132 more than the 54^{th} term.

**Alternative method:**

Let n ^{th} term be 132 more than the 54^{th} term.

n = 54 + 132/2

= 54 + 11

= 65^{th} term