In an A.P. if a = 1, a_n = 20 and S_n = 399, then n is

1. 19

2. 21

3. 38

4. 42

Given a = 1, a_n = 20 and S_n = 399.

We know that

    a_n = a + (n - 1)d
∴ a_n = 1 + (n - 1) × d
⇒ 20 = 1 + (n - 1)d
⇒ (n - 1)d = 20 - 1
⇒ (n - 1)d = 19
⇒ d = $\dfrac{19}{n - 1}$.

The formula for sum of A.P. is given by,

$S$n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore Sn = \dfrac{n}{2}[2 \times 1 + (n - 1) \times d] \\[1em] \Rightarrow 399 = \dfrac{n}{2}[2 + (n - 1) \times \dfrac{19}{n - 1}] \\[1em] \Rightarrow 399 = \dfrac{n}{2}[2 + 19] \\[1em] \Rightarrow \dfrac{n}{2} \times 21 = 399 \\[1em] \Rightarrow n = \dfrac{399 \times 2}{21} \\[1em] \Rightarrow n = \dfrac{798}{21} \\[1em] \Rightarrow n = 38.

Hence, Option 3 is the correct option.