Question

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

• A. 55
• B. 80
• C. 65
• D. 70

Answer 1

The correct option is C

65

Let’s first calculate ${54}^{th}$ of the given AP.

First term $=a=3$

Common difference $=d=15-3=12$

Using formula $a_n=a+(n-1)d$, to find $n^{th}$ term of arithmetic progression, we get

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

$a_{54}=3+53\left(12\right)=3+636=639$

We want to find which term is 132 more than its ${54}^{th}$ term. Let’s suppose it is $n^{th}$ term which is 132 more than ${54}^{th}$ term.

Therefore, we can say that

$a_n=a_{54}+132$

$a_n=a+(n-1)d=3+(n-1)12$

$\Rightarrow 3+(n-1)12=639+132$

$\Rightarrow 3+12n-12=771$

$\Rightarrow 12n-9=771$

$\Rightarrow 12n=780$

$\Rightarrow n=\frac{780}{12}=65$

Therefore, ${65}^{th}$ term of te given AP is 132 more than its ${54}^{th}$ term.