Question

Let $\alpha$ and $\beta$ be the roots of equation $x^2-6x-2=0$. If $a_n={\alpha }^n-{\beta }^n$, for $n⩾1$, then the value of $\frac{a_{10}-2a_8}{2a_9}$ is equal to

• A. $6$
• B. $-6$
• C. $3$
• D. $-3$

Answer 1

The correct option is C $3$

$\because \alpha ,\beta$ are the roots

$\Rightarrow {\alpha }^2-6\alpha -2=0\Rightarrow {\alpha }^2-2=6\alpha$

and ${\beta }^2-6\beta -2=0\Rightarrow {\beta }^2-2=6\beta$

Hence, $\frac{a_{10}-2a_8}{2a_9}=\frac{{\alpha }^{10}-{\beta }^{10}-2({\alpha }^8-{\beta }^8)}{2({\alpha }^9-{\beta }^9)}$

$=\frac{{\alpha }^8({\alpha }^2-2)-{\beta }^8({\beta }^2-2)}{2({\alpha }^9-{\beta }^9)}=\frac{{\alpha }^8.6\alpha -{\beta }^8.6\beta }{2({\alpha }^9-{\beta }^9)}=\frac{6({\alpha }^9-{\beta }^9)}{2({\alpha }^9-{\beta }^9)}=3$