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

Given:

Given A.P. is $3, 15, 27, 39, ….$

To do:

We have to find which term of the given A.P. will be 132 more than its 54th term.

Solution:

Here,

$a_1=3, a_2=15, a_3=27$

Common difference $d=a_2-a_1=15-3=12$

We know that,

nth term $a_n=a+(n-1)d$

Therefore,

$a_{54}=3+(54-1)(12)$

$=3+53(12)$

$=3+636$

$=639$

132 more than the 54th term $=132+639=771$

This implies,

$a_{n}=3+(n-1)12$

$771=3+12n-12$

$12n=771+9$

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

$n=65$

Hence, 65th term is 132 more than the 54th term.