The sum of first n terms of a certain series is given as 2n2-3n . Show that the series is an A.P.

Solution :

Given that :

tn = 2n2 - 3n

Sn = (n/2) [2a + (n - 1)d]

Sn = (n/2) [2(-1) + (n - 1)3]

= (n/2) [-2 + 3n - 3]

= (n/2) [3n - 5]