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

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