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

Answer: Option D

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