Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?

In the given problem, let us first find the 21st term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here,

First term (a) = 3

Common difference of the A.P. $(d)=15-3=12$

Now, as we know,

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

So, for $21^{\text {st }}$ term $(n=21)$,

$a_{21}=3+(21-1)(12)$

$=3+20(12)$

$=3+240$

$=243$

Let us take the term which is 120 more than the 21st term as an. So,

$a_{n}=120+a_{21}$

$=120+243$

$=363$

Also, $a_{n}=a+(n-1) d$

$363=3+(n-1) 12$

$363=3+12 n-12$

$363=-9+12 n$

$363+9=12 n$

Further simplifying, we get,

$372=12 n$

$n=\frac{370}{12}$

$n=31$

Therefore, the $31^{\text {t }}$ term of the given A.P. is 120 more than the $21^{\text {st }}$ term.