[JEE Advanced 2011] Let $\alpha>\beta$ be the roots of $x^2-6x-2=0$ and $a_n=\alpha^n-\beta^n$ ($n\ge1$). Then $\dfrac{a_{10}-2a_8}{2a_9}$ is
Let $\alpha>\beta$ be the roots of $x^2-6x-2=0$ and $a_n=\alpha^n-\beta^n$ ($n\ge1$). Then $\dfrac{a_{10}-2a_8}{2a_9}$ is
(a) $1$
(b) $2$
(c) $3$
(d) $4$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (c) $3$
From $x^2=6x+2$, $a_n=6a_{n-1}+2a_{n-2}$, so $a_{10}-2a_8=6a_9$ and the ratio is $3$.
JEE Advanced 2011 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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