Solution:
Given AP is 3, 15, 27, 39, …
First term, a = 3
Common difference, d = a2 − a1 = 15 − 3 = 12
We know that,
an = a + (n − 1)d
Therefore,
a54 = a + (54 − 1) d
⇒ 3 + (53)(12)
⇒ 3 + 636 = 639
a54 = 639 + 132 = 771
We have to find the term of this AP which is 132 more than a54, i.e., 771.
Let 771 be the nth term of the AP.
an = a + (n − 1) d
771 = 3 + (n − 1)12
768 = (n −1)12
(n −1) = 64
n = 65
Therefore, the 65th term was 132 more than the 54th term.
Alternative method:
Let n th term be 132 more than the 54th term.
n = 54 + 132/2
= 54 + 11
= 65th term