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Answer :
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.