Question

Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36^th term.

Answer 1

In the given problem, let us first find the 36^st term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (a) = 9

Common difference of the A.P. (d) =3

Now, as we know,

So, for 36^th term (n = 36),

Let us take the term which is 39 more than the 36^th term as a_n. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 39 more than the 36^th term