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(D) Given that $\alpha, \beta, \gamma$ are in $G.P.$,we can write $\beta = \alpha r$ and $\gamma = \alpha r^2$ for some common ratio $r 
eq 1$.Substi

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$\beta\gamma$

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(D) Given that $\alpha, \beta, \gamma$ are in $G.P.$,we can write $\beta = \alpha r$ and $\gamma = \alpha r^2$ for some common ratio $r 
eq 1$.Substituting these into the first equation: $\alpha x^2 + 2(\alpha r)x + \alpha r^2 = 0$.Since $\alpha 
eq 0$,we divide by $\alpha$: $x^2 + 2rx + r^2 = 0$,which is $(x + r)^2 = 0$. Thus,the root is $x = -r$.Since this is a common root with $x^2 + x - 1 = 0$,we substitute $x = -r$ into the second equation:$(-r)^2 + (-r) - 1 = 0 \implies r^2 - r - 1 = 0$.From the $G.P.$ properties,$\beta^2 = \alpha\gamma$. Also,$\beta = \alpha r$ and $\gamma = \alpha r^2$.We need to evaluate $\alpha(\beta + \gamma) = \alpha(\alpha r + \alpha r^2) = \alpha^2 r(1 + r)$.Since $r^2 = r + 1$,we have $r^2 - r = 1$. Also,$\beta\gamma = (\alpha r)(\alpha r^2) = \alpha^2 r^3 = \alpha^2 r(r^2) = \alpha^2 r(r + 1) = \alpha^2(r^2 + r)$.Comparing the coefficients of the two equations $\alpha x^2 + 2\beta x + \gamma = 0$ and $x^2 + x - 1 = 0$,for them to have common roots,the ratio of coefficients must be equal: $\frac{\alpha}{1} = \frac{2\beta}{1} = \frac{\gamma}{-1} = k$.Thus,$\alpha = k, \beta = k/2, \gamma = -k$.Since they are in $G.P.$,$\beta^2 = \alpha\gamma \implies (k/2)^2 = k(-k) \implies k^2/4 = -k^2$. This implies $k=0$,which contradicts the non-constant $G.P.$ condition.Re-evaluating: The condition for a common root in $ax^2+bx+c=0$ and $dx^2+ex+f=0$ is $(af-cd)^2 = (ae-bd)(bf-ce)$.Substituting $\alpha, 2\beta, \gamma$ and $1, 1, -1$: $(\alpha(-1) - \gamma(1))^2 = (\alpha(1) - 2\beta(1))(2\beta(-1) - \gamma(1))$.$(-\alpha - \gamma)^2 = (\alpha - 2\beta)(-2\beta - \gamma) \implies \alpha^2 + 2\alpha\gamma + \gamma^2 = -2\alpha\beta - \alpha\gamma + 4\beta^2 + 2\beta\gamma$.Since $\beta^2 = \alpha\gamma$,$\alpha^2 + 2\alpha\gamma + \gamma^2 = -2\alpha\beta - \beta^2 + 4\beta^2 + 2\beta\gamma = -2\alpha\beta + 3\beta^2 + 2\beta\gamma$.Substituting $\beta^2 = \alpha\gamma$,we get $\alpha^2 + 2\beta^2 + \gamma^2 = -2\alpha\beta + 3\beta^2 + 2\beta\gamma \implies \alpha^2 - \beta^2 + \gamma^2 = -2\alpha\beta + 2\beta\gamma$.Using $\alpha = \beta/r, \gamma = \beta r$,we find $\alpha(\beta + \gamma) = \beta\gamma$.

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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