Question

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

• A. t${}_{65}$
• B. t${}_{64}$
• C. t${}_{60}$
• D. None of these
  

Answer 1

The correct option is A t${}_{65}$

Clearly, the given Sequence is an AP with first term $a=3$ and common difference $d=15-3=12$.
Let $nth$ term be the $132$ more than $54th$ term of the given AP.
$\Rightarrow a_n=a_{54}+132$
$\Rightarrow a+(n-1)d=a+53d+132\Rightarrow (n-1)\times 12=53\times 12+132$
$\Rightarrow n-1=53+11$
$\Rightarrow n=65$
Hence, $a_{65}$ is $132$ more than $a_{54}$