Question

Let \(\alpha, \beta\) be the roots of the equation \(x^{2}-6 x-2=0\) with \(\alpha>\beta\). If \(a_{n}=\alpha^{n}-\beta^{n}\) for \(n \geq 1\), then the value of \(\frac{a_{10}-2 a_{8}}{2 a_{9}}\) is
(A) 1
(B) 2
(C) 3
(D) 4

Answer 1

Ans: (C)
Hint: \(x^{2}-6 x-2=0\)
$$
\begin{aligned}
&x^{n}-6 x^{n-1}-2 x^{n-2}=0 \\
&\Rightarrow x^{n}-2 x^{n-2}=6 x^{n-1} \\
&\text { for } n=10
\end{aligned}
$$
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JEE - 2021 (Answers \& Hint)
$$
\begin{aligned}
&x^{10}-2 x^{8}=6 x^{9} \\
&\alpha^{10}-2 \alpha^{8}=6 \alpha^{9} \\
&\beta^{10}-2 \beta^{8}=6 \beta^{9} \\
&\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^{8}-\beta^{8}\right)=6\left(\alpha^{9}-\beta^{9}\right) \\
&\Rightarrow a_{10}-2 a_{8}=6 a_{9} \\
&\Rightarrow \frac{a_{10}-2 a_{8}}{2 a_{9}}=3
\end{aligned}
$$

Answer 2

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Answer 3

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