The number of terms of A.P. 3, 7, 11, 15, ....... to be taken so that sum is 406 is

The A.P. is 3, 7, 11, 15, ......
Where a = 3, d = 7 - 3 = 4 and sum S_n = 406
$$\eqalign{ & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & \Rightarrow 406 = \frac{n}{2}\left[ {2 \times 3 \times \left( {n - 1} \right) \times 4} \right] \cr & \Rightarrow 812 = n\left( {6 + 4n - 4} \right) \cr & \Rightarrow 812 = n\left( {4n + 2} \right) \cr & \Rightarrow 4{n^2} + 2n - 812 = 0 \cr & \Rightarrow 2{n^2} + n - 406 = 0 \cr & \Rightarrow 2{n^2} + 29n - 28n - 406 = 0 \cr & \Rightarrow n\left( {2n + 29} \right) - 14\left( {2n + 29} \right) = 0 \cr & \Rightarrow \left( {2n + 29} \right)\left( {n - 14} \right) = 0 \cr & \therefore n = 14\,{\text{or}}\,\frac{{ - 29}}{2} \cr & {\text{But}}\,n = \frac{{ - 29}}{2}\,{\text{is}}\,{\text{not}}\,{\text{possible}} \cr} $$