Question

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

Answer 1

In the given problem, let us first find the 41^st 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) =6

Now, as we know,

So, for 41^st term (n = 41),

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

Also,

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41^st term.