Find the term of the arithmetic progression 9, 12, 15, 18, ... which is

In the given problem, let us first find the 36st term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (a) = 9

Common difference of the A.P. $(d)=12-9=3$

Now, as we know,

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

So, for 36th term (n = 36),

$a_{36}=9+(36-1)(3)$

$=9+35(3)$

$=9+105$

$=114$

Let us take the term which is 39 more than the 36th term as an. So,

$a_{n}=39+a_{36}$

$=39+114$

$=153$

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

$153=9+(n-1) 3$

$153=9+3 n-3$

$153=6+3 n$

$153-6=3 n$

Further simplifying, we get,

$147=3 n$

$n=\frac{147}{3}$

$n=49$

Therefore, the $49^{\text {th }}$ term of the given A.P. is 39 more than the $36^{\text {th }}$ term