Question

which term of the A.P 8, 14, 20, 26 ,....will be 72 more than its 41st term ?​

Answer 1

Given,

AP: 8, 14, 20, 26,...

To find,

The term 72 more than its 41st term.

Solution,

This numerical problem can be solved simply by following the below steps.

From the given AP, it can be observed that,

The first term, a = 8,

The common difference, d = 14-8 = 6.

Now, let the term which is 72 more than the 41st term be $a_n$.

So, from the given condition,

$a_n=a_{41}+72$.

Since, the $n^{th}$ term of an AP $a_n$ is given by,

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

Substituting this for n and n = 41 in the previous relation,

$[a+(n-1)d]=[a+(41-1)d]+72$

Now, substituting a, d in this equation, we get,

$[8+(n-1)6]=[8+(41-1)6]+72$

⇒ $[8+(n-1)6]=[8+40*6]+72$

⇒ $[8+(n-1)6]=248+72$

⇒ $8+(n-1)6=320$

⇒ $n-1=\frac{312}{6}$

⇒ $n=52+1$

⇒ $n=53.$

Therefore, the term which is 72 more than the 41st term of the given AP will be the 53rd.