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


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