Answer: (4)

Solution

$\because \alpha, \beta, \gamma$ are three consecutive terms of a non-constant G.P.

$\therefore \beta^2 = \alpha\gamma$

So roots of the equation $\alpha x^2 + 2\beta x + \gamma = 0$ are

$\frac{-2\beta \pm 2\sqrt{\beta^2 - \alpha\gamma}}{2\alpha} = \frac{-\beta}{\alpha}$

$\therefore \alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$ have a common root.

$\therefore$ this root satisfy the equation $x^2 + x - 1 = 0$

$\beta^2 - \alpha\beta - \alpha^2 = 0 \Rightarrow \alpha\gamma - \alpha\beta - \alpha^2 = 0 \Rightarrow \alpha + \beta = \gamma$

Now, $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$

$= \alpha\beta + \beta^2 = (\alpha + \beta)\beta = \beta\gamma$