Given,

AP is 3, 15, 27, 34, ……. 

First term of AP is a = 3 and common difference of AP is d = a₂ – a₁ 

= 15 – 3 

= 12. 

54^th term of AP is a₅₄ = a + (54 – 1)d 

(∵ a_n = a + (n – 1) d and n = 54) 

= 3 + 53 × 12 

= 3 + 636 

= 639. 

(∵ a = 3 & d = 12) 

Let n^th term of AP is 132 more than its 54^th term. 

i.e.,

a_n = 132 + a₅₄ 

= 132 + 639 = 771 

⇒ 3 + (n – 1) 12 = 771 

(∵ a = 3 & d = 12 and a_n = a + (n – 1) d)

⇒ 12(n − 1) = 771 − 3 = 768 

⇒ n − 1 = \(\frac{768 }{12}\) = 64 

⇒ n = 64 + 1 = 65.

Hence,

65^th term of AP is 132 more than its 54^th term.

Given a = 3, d = 12, n = 54.
We know that an=a+(n-1) * d   ---------- (1)

                           = 3 + (54-1) * 12
                           = 3 + 53 * 12
                          =639

                  a54 = 639.
 

Given that the term is 132 more than its 54th term = 639 + 132 = 771.
on substituting in (1), we get

771 = 3 + (n-1) * 12
768 = (n-1) * 12
768/12 = n-1
n-1 = 64
n = 65.
 

Therefore the 65th term is 132 more than its 54th term.Given a = 3, d = 12, n = 54.

We know that an=a+(n-1) * d   ---------- (1)

                            = 3 + (54-1) * 12

                            = 3 + 53 * 12

                            = 639.

                     a54 = 639.
 

Given that the term is 132 more than its 54th term = 639 + 132 = 771.

on substituting in (1), we get

771 = 3 + (n-1) * 12

768 = (n-1) * 12

768/12 = n-1

n-1 = 64

n = 65.

Therefore the 65th term is 132 more than its 54th term.