The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
Let ‘a’ be the first term and ‘d’ be the common difference
Sn= 3n2 + 6n
Firsdt term, a, S1 = 3(1)2 + 6(1)
= 3 + 6 = 9
S2 = 3(2)2 + 6(2)
a + a + d = 12 + 12
9 + 9 + d = 24
18 + d = 24
d = 6
Therefore, an = a + (n – 1) d
= 9 + (n – 1) 6
= 6n + 3