If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 term is

Solution (By Examveda Team)

$$\eqalign{ & {\text{In}}\,{\text{an}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & a = 2\,{\text{and}}\,d = 4,\,n = 40 \cr & \therefore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{{40}}{2}\left[ {2 \times 2 + \left( {40 - 1} \right) \times 4} \right] \cr & = 20\left[ {4 + 39 \times 4} \right] \cr & = 20 \times \left( {4 + 156} \right) \cr & = 20 \times 160 \cr & = 3200 \cr} $$