Question

Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$, with $\alpha >\beta$. lf $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1$, then the value of $\frac{a_{10}-2a_8}{2a_9}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

Since $\alpha$ and $\beta$ are the roots(solutions) of the equation
$x^2-6x-2=0$
So, it will satisfy the equation
${\alpha }^2-6\alpha -2=0$ .....(1)
${\beta }^2-6\beta -2=0$ .....(2)
Now, we have to find the value of $\frac{a_{10}-2a_8}{2a_9}$
Given, $a_n={\alpha }^n-{\beta }^n$
So, $a_{10}={\alpha }^{10}-{\beta }^{10}$ ......(3)
So, to get the values of ${\alpha }^{10}$ and ${\beta }^{10}$ , multiplying equation (1) and (2) by ${\alpha }^8$ and ${\beta }^8$ respectively.
${\alpha }^{10}-6{\alpha }^9-2{\alpha }^8=0$
or , ${\alpha }^{10}=6{\alpha }^9+2{\alpha }^8$ ......(4)
${\beta }^{10}-6{\beta }^9-2{\beta }^8=0$
or, ${\beta }^{10}=6{\beta }^9+2{\beta }^8$ ......(5)
So, using equation(4) and (5) in (3), we get
$a_{10}=6{\alpha }^9+2{\alpha }^8-(6{\beta }^9+2{\beta }^8)$
$a_{10}=6{\alpha }^9+2{\alpha }^8-6{\beta }^9-2{\beta }^8$
$=6({\alpha }^9-{\beta }^9)+2({\alpha }^8-{\beta }^8)$
$a_{10}=6a_9+2a_8$
$a_{10}-2a_8=6a_9$
$\frac{a_{10}-2a_8}{2a_9}=3$