Question

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

Answer 1

Determine the term that is $132$ terms longer than the$54th$ term.

Given,

$A.P.is3,15,27,39,\dots$

First term,$a=3$, $n=54$, common difference $d=15-3=27-15=12$

$$\begin{gathered} a_{{54}^{th}}=a+(n–1)d \\ =3+(54–1)12 \end{gathered}$$

$=3+53\times 12$

$=3+636$

$=639$

Let $n^{th}$ term is 132 more than $54th$ term

$132+639=771=a_n$

$771=3+(n–1)12$

$\Rightarrow$$\frac{768}{12}+1=n$

$\Rightarrow$ $64+1=n$

$\Rightarrow$ $n=65$

Hence , the ${65}^{th}$ term is $132$ more than ${54}^{th}$ term.