#### Question

In an A.P., the sum of first n terms is `(3n^2)/2 + 13/2 n`. Find its 25^{th} term.

#### Solution

Here, the sum of first *n* terms is given by the expression,

S_n = `(3n^2)/2 + 13/2 n`

We need to find the 25^{th} term of the A.P.

So we know that the n^{th}term of an A.P. is given by,

`a_n = S_n- S_(n - 1)`

So `a_25 = S_25 - S_24` ....(1)

So, using the expression given for the sum of n terms, we find the sum of 25 terms (S_{25}) and the sum of 24 terms (S_{24}). We get,

`S_25 = (3(25)^2)/2 + 13/2 (25)`

`= (3(25)^2)/2 + 13/2 (25)`

`= (3(625))/2 + (13(25))/2`

`= 1875/2 = 325/2`

= 2200/2

= 1100

Similarly

`S_24 = (3(24)^2)/2 + 13/2 (24)`

`= (3(576))/2 + (13(24))/2`

`= 1728/2 + 312/2`

`= 2040/2`

=1020

Now, using the above values in (1),

`a_25 = S_25 - S_24`

= 1100 - 1020

= 80

Therefore `a_25 = 80`