# Which term of the arithmetic progression $8, 14, 20, 26, â€¦$ will be 72 more than its 41st term?

Given:

Given A.P. is $8, 14, 20, 26, …$

To do:

We have to find which term of the given A.P. will be 72 more than its 41st term.

Solution:

Here,

$a_1=8, a_2=14, a_3=20$

Common difference $d=a_2-a_1=14-8=6$

We know that,

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

Therefore,

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

$=8+240$

$=248$

72 more than the 41st term $=72+248=320$

This implies,

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

$320=8+6n-6$

$6n=320-2$

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

$n=53$

Hence, 53rd term is 72 more than the 41st term.â€Š

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