Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2-6x-2=0.$ If $a_n=\alpha^n-\beta^n,$ for $n\ge 1$ , then the value of $\frac{a_{10}-2a_{8}}{2 a_9}$ is

• A. 6
• B. -6
• C. 3
• D. -3

Answer 1

Answer: (C)

$ x^{2}-6 x-2=0<_{\beta}^{\alpha} $
$\Rightarrow \alpha^{2}-6 \alpha-2=0 $
$ \beta^{2}-6 \beta-2=0 \,\,\,\,\ldots . .(1) $
$a_{n}=\alpha_{n}-\beta_{n} n \geq 1$
$a_{40}-29_{8}=\alpha^{10}-\beta^{10}-2 \alpha^{8}+2 \beta^{8}$
$=\alpha^{8}\left(a^{2}-2\right)-b^{8}\left(b^{2}-2\right)$
$=\alpha^{8}(6 \alpha)-\beta^{8}(6 \beta) \,\,\,\,\,\,\,$ (using (1)
$=6 \alpha^{9}-6 \beta^{9}$
$=6 a_{9}$
now
$\frac{a_{10}-2 a_{8}}{2 a_{9}}=\frac{6 a_{9}}{2 a_{9}}=3$