$S_1 = 3(1)^2 + 1 = 4$. Since $S_1 = a_1$, the first term is 4.
(D) 4
The first term of any AP equals $S_1$ (sum of first 1 term). Substitute $n = 1$ in $S_n = 3n^2 + n$: $S_1 = 3 + 1 = 4$. You can verify: $a_2 = S_2 - S_1 = (12+2)-(3+1) = 10$, and $a_2 - a_1 = 10 - 4 = 6$ ✓ (matches given $d = 6$).