Let $\alpha =a,\beta =ar,\gamma =a{r}^2,\delta =a{r}^3$

$\alpha +\beta =3\Rightarrow a+ar=3$  ...........$\left(1\right)$

$\gamma +\delta =6\Rightarrow a{r}^2+a{r}^3=6$   .........$\left(2\right)$

By $\left(1\right)$ and $\left(2\right)$ $\Rightarrow \frac{a{r}^2\left(1+r\right)}{a\left(1+r\right)}=\frac{6}{3}$

$\Rightarrow r^2=2$

$\therefore$  $\frac{2q+p}{2q-p}=\frac{2\gamma \delta +\alpha \beta }{2\gamma \delta -\alpha \beta }=\frac{2a^2{r}^5+a^{2}r}{2a^2{r}^5-a^{2}r}=\frac{2r^4+1}{2r^4-1}=\frac{8+1}{8-1}=\frac{9}{7}$