If an A.P. has a = 1, t_n = 20 and s_n = 399, then value of n is :

Solution (By Examveda Team)

$$\eqalign{ & {S_n} = \frac{1}{2}\left( {a + l} \right) \times n \cr & \Rightarrow 399 = \left( {1 + 20} \right) \times \frac{n}{2} \cr & \Rightarrow 399 \times 2 = 21 \times n \cr & \Rightarrow n = 399 \times \frac{2}{{21}} \cr & \Rightarrow n = 19 \times 2 \cr & \Rightarrow n = 38 \cr} $$