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,
an = 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 a54, i.e.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
As a result, 65th term was 132 more than 54th term.
Another method is;
Let the nth term should be 132 more than the 54th term.
n = 54 + 132/2 = 54 + 11 = 65th term