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(D) Since $\alpha, \beta, \gamma$ are in $G.P.$,we have $\beta^2 = \alpha\gamma$. Given that the equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x

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$\beta\gamma$

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(D) Since $\alpha, \beta, \gamma$ are in $G.P.$,we have $\beta^2 = \alpha\gamma$. Given that the equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$ have a common root,and the coefficients are proportional,both roots must be common. Thus,$\frac{\alpha}{1} = \frac{2\beta}{1} = \frac{\gamma}{-1} = k$. This implies $\alpha = k$,$\beta = \frac{k}{2}$,and $\gamma = -k$. Substituting these into the expression $\alpha(\beta + \gamma)$: $\alpha(\beta + \gamma) = k(\frac{k}{2} - k) = k(-\frac{k}{2}) = -\frac{k^2}{2}$. Also,$\beta\gamma = (\frac{k}{2})(-k) = -\frac{k^2}{2}$. Therefore,$\alpha(\beta + \gamma) = \beta\gamma$.

If $\alpha, \beta$ and $\gamma$ are three consecutive terms of a non-constant $G.P.$ such that the equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$ have a common root,then $\alpha(\beta + \gamma)$ is equal to

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