Question

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

• A. ${64}^{th}$
• B. ${56}^{th}$
• C. ${65}^{th}$
• D. ${67}^{th}$

Answer 1

The correct option is C ${65}^{th}$

$a=3$
$d=15-3=12$
Let the required term be the ​$n^{th}$ term.
$a_n=a+(n-1)d$
$a_{54}+132=(32+53\times 12)+132$
$=32+636+132=800=a_n$
$800=32+(n-1)12$
$768=(n-1)12n-1=\frac{768}{12}=64n=64+1=65$
Thus, ${65}^{th}$ term is 132 more than ${54}^{th}$ term.