Question

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

• A. $60$
• B. $62$
• C. $65$
• D. $63$

Answer 1

The correct option is C $65$

Given series is $3,15,27,\mathrm{39.....}$
We know,

$n^{th}$ term $a_n=a+(n-1)d$

$a=$ First term

$d=$ Common difference between consecutive terms

$n=$ number of terms

$a_n=n^{th}$ term

Here $a=3$ and $d=15-3=27-15=12$

The $54$th term of this sequence is

$t_{54}=3+(54-1)12⟹t_{54}=3+636=639$
The term which will be $132$ more than $639$ will be $771$.
So, we need to find which term in the sequence is $771.$
$771=3+(n-1)\left(12\right)⟹771=3+12n-12⟹771=12n-9⟹12n=780orn=65$.
Hence, $771$ is the $65$th term.