Question

Which term of the $AP:3,15,27,39,....$ will be $132$ more than its $54th$ term?

Answer 1

A.P: $3,15,27,39,....$

Common difference, $d=12$.

$\therefore t_n=a+(n-1)d$ where $a=3,d=12$

$\therefore t_{54}=3+(54-1)12$

$=3+12\times 53$

$=3+636$

$=639$

$t_m=639+132=771$

$\therefore 3+(m-1)12=771$

$m-1=\frac{771-3}{12}$

$=64$

$\therefore m=65$