Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = [2a + (n − 1)d]
It is given that sum of the first n terms of an A.P. is 3n2 + 6n.
∴ First term = a = S1 = 3(1)2 + 6(1) = 9.
Sum of first two terms = S2 = 3(2)2 + 6(2) = 24.
∴ Second term = S2 − S1 = 24 − 9 = 15.
∴ Common difference = d = Second term − First term
= 15 − 9 = 6
Also, nth term = an = a + (n − 1)d
⇒ an = 9 + (n − 1)6
⇒ an = 9 + 6n − 6
⇒ an = 3 + 6n
Thus, nth term of this A.P. is 3 + 6n.