The sum of 25 terms of the A.P., $-\dfrac{2}{3}, -\dfrac{2}{3}, -\dfrac{2}{3}, …$ is

1. 0

2. $-\dfrac{2}{3}$

3. $-\dfrac{50}{3}$

4. -50

The above series is an A.P. with first term = $-\dfrac{2}{3}$ and common difference = d = $-\dfrac{2}{3} - \Big(-\dfrac{2}{3}\Big) = 0$

We know that,

$S$n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore S{25} = \dfrac{25}{2}\Big[2 \times -\dfrac{2}{3} + (25 - 1) \times 0\Big] \\[1em] = \dfrac{25}{2}\Big[-\dfrac{4}{3} + 24 \times 0\Big] \\[1em] = \dfrac{25}{2} \times -\dfrac{4}{3} \\[1em] = -\dfrac{50}{3}.

Hence, Option 3 is the correct option.