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

Here, a = 3, d = 15 – 3 = 12

54th term can be given as follows:

a₅₄ = a + 53d

= 3 + 53 x 12

= 3 + 636 = 639

So, the required term = 639 + 132 = 771

771 = a + (n – 1)d

771 = 3 + (n -1)12

 (n – 1)12 = 771 – 3 = 768

n – 1 = 64

n = 65

Thus, the required term is 65th term