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

(65) Given $A.P.$ is $3, 15, 27, 39, \ldots$
First term $a = 3$.
Common difference $d = 15 - 3 = 12$.
The $54^{th}$ term is given by $a_{54} = a + (54 - 1)d$.
$a_{54} = 3 + 53 \times 12 = 3 + 636 = 639$.
We need to find the term $a_n$ such that $a_n = a_{54} + 132$.
$a_n = 639 + 132 = 771$.
Using the formula $a_n = a + (n - 1)d$:
$771 = 3 + (n - 1)12$.
$768 = (n - 1)12$.
$n - 1 = \frac{768}{12} = 64$.
$n = 65$.
Alternatively,the difference between the $n^{th}$ term and the $54^{th}$ term is $(n - 54)d = 132$.
$(n - 54)12 = 132$.
$n - 54 = 11$.
$n = 65$.