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

Solution:

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.