Question

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

A) 1

B) 2

C) 3

D) 4

Answer 1

Hint: Since it is given that $\alpha$ and $\beta$ are the roots of the $x^2-6x-2=0$, then we will put the value of x as $\alpha$ and $\beta$ in the given equation and find the value.

Complete step-by-step answer:

It is given that $\alpha$ and $\beta$ are the roots of the $x^2-6x-2=0$.

Hence, ${\alpha }^2-6\alpha -2=0$

${\alpha }^2=6\alpha +2$

Multiply ${\alpha }^8$ to both sides of the given equation, we get

${\alpha }^2\times {\alpha }^8={\alpha }^8(6\alpha +2)$

${\alpha }^{10}=6{\alpha }^9+2{\alpha }^8$ (Since bases are same, we can add the powers) ……………… (1)

Similarly, we will put the value of x as $\beta$, we get

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

${\beta }^2=6\beta +2$

Multiply $${\beta }^8$$ to both sides of the given equation, we get

${\beta }^2\times {\beta }^8={\beta }^8(6\beta +2)$

${\beta }^{10}=6{\beta }^9+2{\beta }^8$ (Since bases are same, we can add the powers) ……………… (2)

Now, we have to find the value of ${\displaystyle \frac{a_{10}-2a_8}{2a_9}}$ and it is given that $a_n={\alpha }^n-{\beta }^n$ (where $\alpha$>$\beta$)

Now, we will put n = 10 in $a_n={\alpha }^n-{\beta }^n$, we get

$a_{10}={\alpha }^{10}-{\beta }^{10}$, $a_8={\alpha }^8-{\beta }^8$

Now, we have 

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

Put the value of (1) and (2), we get

${\displaystyle \frac{a_{10}-2a_8}{2a_9}}$= ${\displaystyle \frac{6{\alpha }^9+2{\alpha }^8-6{\beta }^9-2{\beta }^8-2{\alpha }^8+2{\beta }^8}{2a_9}}$

= ${\displaystyle \frac{6({\alpha }^9-{\beta }^9)}{2a_9}}$

= ${\displaystyle \frac{6a_9}{2a_9}}$= 3

Therefore, option (C) is correct.

Note: In this question, we are given an equation, of which $\alpha$ and $\beta$ are the two roots, so we have put in the value of x as $\alpha$ and $\beta$ and solve it then we will find the value of a10 (since the value of an is also given as $a_n={\alpha }^n-{\beta }^n$ for $n⩾1,$). Put all the values in ${\displaystyle \frac{a_{10}-2a_8}{2a_9}}$ to get the desired result.