The product of first three terms of a G.P. is −1 and the common ratio is $-\dfrac{3}{4}$. The sum of these three terms is :

1. $\dfrac{12}{13}$

2. $\dfrac{11}{13}$

3. $\dfrac{11}{12}$

4. $\dfrac{13}{12}$

Let the first three terms of the G.P. be $\dfrac{a}{r}$, a, ar.

The product of the three terms is given as -1

⇒ $\dfrac{a}{r}$ × a × ar = -1

⇒ a³ = -1

⇒ a = $\sqrt[3]{-1}$

⇒ a = -1.

Substitute a = -1 and r = $\dfrac{-3}{4}$, we get :

⇒ $\dfrac{a}{r} = \dfrac{-1}{\dfrac{-3}{4}} = \dfrac{4}{3}$,

⇒ ar = $(-1) \times \Big(-\dfrac{3}{4}\Big) = \dfrac{3}{4}$.

The sum of the three terms is :

$\Rightarrow \dfrac{4}{3} + (-1) + \dfrac{3}{4} \\[1em] = \dfrac{16 - 12 + 9}{12} \\[1em] = \dfrac{25 - 12}{12} \\[1em] = \dfrac{13}{12}.$

Hence, option 4 is the correct option.