Question

Which term of the A.P. $3,15,27,39,\dots$ will be $132$ more than its ${54}^{th}$ term?

Answer 1

Given A.P. is $3,15,27,39,\dots$

First term, $a=3$

Common difference, $d=a_2-a_1=15-3=12$

We know that the $n^{th}$ term of an A.P is given by $a_n=a+(n-1)d$

$a_{54}=a+(54-1)d$

$=3+\left(53\right)\left(12\right)$

$=3+636=639$

so, $132$ more than ${54}^{th}$ term will be $132+639=771$

We have to find the term of this A.P. which is $771.$

Let $n^{th}$ term be $771.$

$a_n=a+(n-1)d$

$\Rightarrow 771=3+(n-1)12$

$\Rightarrow 768=(n-1)12$

$\Rightarrow (n-1)=64$

$\Rightarrow n=65$

Therefore, ${65}^{th}$ term will be $132$ more than ${54}^{th}$ term of the given A.P.​