If the first term of an A.P. is -5 and the common difference is 2, then the sum of its first 6 terms is

1. 0

2. 5

3. 6

4. 15

Given, a = -5 and d = 2.

We know that,

$S$n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S6 = \dfrac{6}{2}[2 \times -5 + (6 - 1) \times 2] \\[1em] = 3[-10 + 5 \times 2] \\[1em] = 3[-10 + 10] \\[1em] = 3 \times 0 \\[1em] = 0.

Hence, Option 1 is the correct option.