If α, β and γ are three consecutive terms of a non-constant G.P.

Question:

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 :

  1. $\beta \gamma$

  2. 0

  3. $\alpha \gamma$

  4. $\alpha \beta$


Correct Option: 1

Solution:

$\alpha x^{2}+2 \beta x+\gamma=0$

Let $\beta=\alpha t, \gamma=\alpha t^{2}$

$\therefore \alpha x^{2}+2 \alpha t x+\alpha t^{2}=0$

$\Rightarrow x^{2}+2 t x+t^{2}=0$

$\Rightarrow(x+t)^{2}=0$

$\Rightarrow x=-t$

it must be root of equation $x^{2}+x-1=0$

$\therefore \mathrm{t}^{2}-\mathrm{t}-1=0$  ................(1)

Now

$\alpha(\beta+\gamma)=\alpha^{2}\left(t+t^{2}\right)$

Option $1 \beta \gamma=\alpha t \cdot \alpha t^{2}=\alpha^{2} t^{3}=a^{2}\left(t^{2}+t\right)$

(from equation 1)

 

 

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