Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $\mathrm{x}^{2}-6 \mathrm{x}+\mathrm{q}=0$. If $\alpha$, $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 \mathrm{q}+\mathrm{p}):(2 \mathrm{q}-\mathrm{p})$ is :
Correct Option: , 3
$x^{2}-3 x+p=0<\begin{aligned}&\alpha \\&\beta\end{aligned}$
$\alpha, \beta, \gamma, \delta$ in G.P.
$\alpha+\alpha r=3 \quad \ldots(1)$
$x^{2}-6 x+q=0<\begin{aligned}&\gamma \\&\delta\end{aligned}$
$\alpha r^{2}+\alpha r^{3}=6$ ..........(2)
$(2) \div(1)$
r^{2}=2
So, $\frac{2 \mathrm{q}+\mathrm{p}}{2 \mathrm{q}-\mathrm{p}}=\frac{2 \mathrm{r}^{5}+\mathrm{r}}{2 \mathrm{r}^{5}-\mathrm{r}}=\frac{2 \mathrm{r}^{4}+1}{2 \mathrm{r}^{4}-1}=\frac{9}{7}$
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