Let α and β be the roots of x^2-6 x-2=0 with α>β . If a_n=α^n-β^n...

$\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^8-\beta^8\right)}{2\left(\alpha^9-\beta^9\right)}$ $=\frac{\alpha^8\left(\alpha^2-2\right)-\beta^8\left(\beta^2-2\right)}{2\left(\alpha^9-\beta^9\right)}$ \left[\because \alpha\right.$is root of$x^2-6 x-2=0$\left.\alpha^2-2=6 \alpha\right]$ Also, $\beta$ is root of $x^2-6 x-2=0$ \begin{aligned} & \Rightarrow \quad \beta^2-2=6 \beta \\ & =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3 \end{aligned}