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

Solution(By Examveda Team)

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