Question

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

Answer 1

Step 1: Identify the first term and common difference of the arithmetic progression

Given that $3,15,27,39,.....$ is in A.P.

$3$ is the first term of the given A.P denoted by $a$

Let $d$ be the common difference between two successive terms of the A.P

$\Rightarrow$$d=15-3=27-15$

$\Rightarrow$$d=12$

Step 2: Write the expression for the $54th$ term of the A.P

The $nth$ term of the A.P is given as

$a_n=a+\left(n-1\right)d$

$\Rightarrow$$a_{54}=3+\left(53\right)\times 12$

$\Rightarrow$$a_{54}=639$

Step 3: Use the given condition to find the required term

It is known that the required term is $132$ more than the $54th$ term

Let the required term be the $nth$term. Then,

$a_n=a_{54}+132$

Substituting the required values we get

$\Rightarrow$ $a+\left(n-1\right)d=639+132$

$\Rightarrow$$3+\left(n-1\right)\times 12=771$

$\Rightarrow$ $\left(n-1\right)\times 12=768$

$\Rightarrow$ $\left(n-1\right)=64$

$\Rightarrow$ $n=65$

Hence, the $65th$ term of the A.P exceeds the $54th$ term by $132$.