$\begin{array}{l} I f α, β a n d γ a r e t h e c o n \sec u t i v e t e r m s o f a n o n - c o n s \tan t G . P . s u c h t h a t t h e e q u a t i o n α x^{2} + 2 β x + γ = 0 \\ a n d x^{2} + x - 1 = 0 h a v e a c o m m o n r o o t, t h e n α (β + γ) i s e q u a l t o : \end{array}$

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$β γ$

$α γ$

$0$

$α β$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t α = a, β = a r, γ = a r^{2} \\ G i v e n E q u a t i o n s α x^{2} + 2 β x + γ = 0 \\ \Rightarrow a x^{2} + 2 a r x + a r^{2} = 0 \\ \Rightarrow x^{2} + 2 r x + r^{2} = 0 \\ \Rightarrow {(x + r)}^{2} = 0 \\ \Rightarrow x = - r \\ I f m u s t b e r o o t o f x^{2} + x - 1 = 0 \\ \Rightarrow r^{2} - r - 1 = 0 \Rightarrow r^{2} = r + 1 \\ N o w α (β + 1) = a (a r + a r^{2}) \\ \Rightarrow a^{2} (r + r^{2}) \\ f r o m (1) \\ β γ = (a r) (a r^{2}) \\ = a^{2} (r (r + 1)) \\ = a^{2} (r^{2} + r) \end{array}$