Q50 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
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
- A6
- B4
- Correct answerC3
- D2
Solution
$\begin{aligned}
& \alpha^2-6 \alpha-2=0 \Rightarrow \alpha^{10}-2 \alpha^8=6 \alpha^9 ....(i)\\
& \beta^2-6 \beta-2=0 \Rightarrow \beta^{10}-2 \beta^8=6 \beta^9 .....(ii)
\end{aligned}$
$\begin{aligned}
& \text { (i) }-(\text { ii) } \\
& \alpha^{10}-\beta^{10}-2\left(\alpha^8-\beta^8\right)=6 \cdot\left(\alpha^9-\beta^9\right) \\
& \Rightarrow a_{10}-2 a_8=6 a_9 \Rightarrow \frac{a_{10}-2 a_8}{2 a_9}=3 .
\end{aligned}$
Hence, the correct answer is option: 'C'.