Given,
AP is 3, 15, 25, 34, ……… .
First term of AP is a = 3
Common difference of AP is d = a2 – a1 = 15 – 3 = 12
54th term of AP = a54 = a + (54 -1) d
= 3 + 53 × 12
= 3 + 636 = 639.
(∵ a = 3 & d = 12)
Let nth term of AP will be 132 more than its 54th term.
∴ an = 132 + 639 = 771.
∴ a + (n – 1) d = 771
(∵ an = a + (n – 1) d)
⇒ 3 + (n − 1)12 = 771
(∵ a = 3 & d = 12)
⇒ 12(n − 1) = 771 − 3
= 768
⇒ n − 1 = \(\frac{768}{12}\) = 64
⇒ n = 64 + 1
= 65.
Hence,
65th term of given AP will be 132 more than its 54th term.