If the sum of n terms of an A.P. is given by

Let Sn denote sum of n terms of an A.P

Such that Sn = 3n + 2n2

i.e S1 = 3(1) + 2(1)2

S1 = 3 + 2 = 5

where S1 = a (first term only)

S2 = a + (a + d) where a + d represents second term and d is common difference.

$S_{2}=3(2)+2(2)^{2}$

$=6+2(4)$

$=6+8$

$S_{2}=14$

i.e 2a + d = 14

i.e 10 + d = 14

d = 4

i.e common difference of A.P is 4

Hence, the correct answer is option D.