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

Answer: Option C

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