Which term of the AP: 3, 15, 27, 39… will be 132 more than its \(54^{th}\) term?

Lets first calculate \(54^{th}\) of the given AP.

First term = a = 3,
Common difference = d
= 15 – 3 = 12

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(a_{54}\) \)
\( = a + (54 - 1) d \)
\( = 3 + 53 (12) \)
= 3 + 636 = 639

We want to find which term is 132 more than its \(54^{th}\) term.

Let us suppose it is nth term which is 132 more than \(54^{th}\) term.

\(\Rightarrow a_n = a_{54} + 132\)
\(\Rightarrow \) 3 + (n - 1) 12 = 639 + 132
\(\Rightarrow \) 3 + 12n – 12 = 771
\(\Rightarrow \) 12n – 9 = 771
\(\Rightarrow \)12n = 780
\(\Rightarrow \) n = 65

Therefore, 65th term is 132 more than its \(54^{th}\) term.