Which term of the AP 8, 14, 20, 26, ... will be 72 more than its 41st term?

In the given problem, let us first find the 41st term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (a) = 8

Common difference of the A.P. $(d)=14-8=6$

Now, as we know,

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

So, for 41st term (n = 41),

$a_{41}=8+(41-1)(6)$

$=8+40(6)$

$=8+240$

$=248$

Let us take the term which is 72 more than the 41st term as an. So,

$a_{n}=72+a_{41}$

$=72+248$

$=320$

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

$320=8+(n-1) 6$

$320=8+6 n-6$

$320=2+6 n$

$320-2=6 n$

Further simplifying, we get,

$318=6 n$

$n=\frac{318}{6}$

$n=53$

Therefore, the $53^{\text {rd }}$ term of the given A.P. is 72 more than the $41^{\text {st }}$ term.