In an A.P., if a = -5, l = 21 and S = 200, then n is equal to

1. 50

2. 40

3. 32

4. 25

We know that

    l = a + (n - 1)d
∴ 21 = -5 + (n - 1) × d
⇒ (n - 1)d = 21 + 5
⇒ (n - 1)d = 26
⇒ d = $\dfrac{26}{n - 1}$.

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 (-5) + (n - 1) \times \dfrac{26}{n - 1}] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[-10 + (n - 1) \times \dfrac{26}{n - 1}] \\[1em] \Rightarrow 200 = \dfrac{n}{2}[-10 + 26] \\[1em] \Rightarrow \dfrac{n}{2} \times 16 = 200 \\[1em] \Rightarrow 8n = 200 \\[1em] \Rightarrow n = \dfrac{200}{8} \\[1em] \Rightarrow n = 25.

Hence, Option 4 is the correct option.