Question

Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41st 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) = 14 – 8 = 6

Now as we know

`a_n = 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 41^st term as a_n So

`a_n = 72 + a_41`

= 72 + 248

= 320

Also `a_n = a + (n - 1)d`

320 = 8 + (n – 1)6

320 = 8 + 6n – 6

320 = 2 + 6n

320 – 2 = 6n

Further simplifying we get

318 = 6n

`n = 318/6`

n = 53

Therefore, the 53^rd term of the given A.P is 72 more than the 41^st term.