If $\alpha, \beta$ are the roots of the equation $x^2-6 x-2=0$, $\alpha\gt\beta$ and $a_{\mathrm{n}}=\alpha^{\mathrm{n}}-\beta^{\mathrm{n}}, n\gt1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \alpha^2-6 \alpha-2=0 \Rightarrow \alpha^{10}-2 \alpha^8=6 \alpha^9 ....(i)\ & \beta^2-6 \beta-2=0 \Rightarrow \beta^{10}-2 \beta^8=6 \beta^9 .....(ii) \end{aligned}$ $\begin{aligned} & 	ext { (i) }-(	ext { ii) } \ & \alpha^{10}-\beta^{10}-2\left(\alpha^8-\beta^8ight)=6 \cdot\left(\alpha^9-\beta^9ight) \ & \Rightarrow a_{10}-2 a_8=6 a_9 \Rightarrow \frac{a_{10}-2 a_8}{2 a_9}=3 . \end{aligned}$

If $\alpha, \beta$ are the roots of the equation $x^2-6 x-2=0$, $\alpha\gt\beta$ and…

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