n an AP, if a = 1, an = 20 and

(c) $\because$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$399=\frac{n}{2}[2 \times 1+(n-1) d]$

$798=2 n+n(n-1) d$ ...(i)

and $\quad a_{n}=20$

$\Rightarrow \quad a+(n-1) d=20 \quad\left[\because a_{n}=a+(n-1) d\right]$

$\Rightarrow \quad 1+(n-1) d=20 \Rightarrow(n-1) d=19$ ...(ii)

Using Eq. (ii) in Eq. (j), we get

$798=2 n+19 n$

$\Rightarrow \quad 798=21 n$

$\therefore$ $n=\frac{798}{21}=38$