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