Question

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

Answer 1

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

Clearly, its First term $=3$ and, Common difference $=12$.

Let $n^{th}$ term of the A.P. be $132$ more than than its ${54}^{th}$ term.

i.e. $a_n=132+a_{54}$

$⟹a+(n-1)d=132+(a+53d)$

$⟹3+12(n-1)=132+(3+53\times 12)$

$⟹12n-9=771$

$⟹12n=780$

$⟹n=65$

Hence, ${65}^{th}$ term of the given A.P. is $132$ more than its ${54}^{th}$ term.