The sum of the first three terms of a GP is $\frac{13}{12}$ and their product is -1. Find the common ratio and the terms.

Let the three terms of the GP be $\frac{a}{r}$, $a$, and $ar$.

Using the product condition:

$$\left(\frac{a}{r}\right)(a)(ar) = a^3 = -1$$

$$\Rightarrow \quad a = -1$$

Using the sum condition: $$\frac{a}{r} + a + ar = \frac{13}{12}$$

Substitute = -1:

$$-\frac{1}{r} - 1 - r = \frac{13}{12}$$

$$\Rightarrow -12 - 12r - 12r^2 = 13r$$

$$\Rightarrow 12r^2 + 25r + 12 = 0$$

On solving the quadratic eq.,

$$(3r + 4)(4r + 3) = 0$$

$$r = -\frac{4}{3} \text{ or } r = -\frac{3}{4}$$

Case 1: $$r = -\frac{4}{3}$$

$$\frac{a}{r} = \frac{-1}{-\frac{4}{3}} = \frac{3}{4}$$

Terms: $$\frac{3}{4}, -1, \frac{4}{3}$$

Case 2: $$r = -\frac{3}{4}$$

$$\frac{a}{r} = \frac{-1}{-\frac{3}{4}} = \frac{4}{3}$$

Terms: $$\frac{4}{3}, -1, \frac{3}{4}$$