Which term of the arithmetic progression 8, 14, 20, 26, ... will

Question:

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

Solution:

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,

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

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,

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

$=72+248$

 

$=320$

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.

 

 

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