#### Question

Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41^{st} 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_n = a+(n + 1)d`

So for 41^{st} term (n = 41)

`a_(41) = 8 + (41 - 1)(6)`

= 8 + 40(6)

= 8 + 240

= 248

Let us take the ter 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

Futher simplifying weget

318 = 6n

`n = 318/6`

n = 53

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

Is there an error in this question or solution?

Solution Which Term of the A.P. 8, 14, 20, 26, ... Will Be 72 More than Its 41st Term? Concept: nth Term of an AP.