If the sum of n terms of an A.P. be 3n2 + n and its common difference is 6, then its first term is

Answer: Option D

Sum of n terms of an A.P. = 3n² + n
and common difference (d) = 6
Let first term be a, then
$$\begin{array}{rl}& \therefore S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]=3n^2+n\\ & \Rightarrow \frac{n}{2}\left[2a+\left(n-1\right)6\right]=3n^2+n\\ & \Rightarrow 2a+6n-6=\left(3n^2+n\right)\times \frac{2}{n}\\ & \Rightarrow 2a+6n-6=n\frac{\left(3n+1\right)\times 2}{n}\\ & \Rightarrow 2a+6n-6=\left(3n+1\right)2\\ & \Rightarrow 2a+6n-6=6n+2\\ & \Rightarrow 2a=6n+2-6n+6\\ & \Rightarrow 2a=8\\ & \therefore a=\frac{8}{2}\\ & =4\end{array}$$