IIT JEE Question

We know that $\alpha +\beta =6$ and $\alpha \beta =-2$

Given $a_n = \alpha^n - \beta^n \quad \forall n \in \mathbb{N}$

Now $\frac { { a }_{ 10 }-2{ a }_{ 8 } }{ 2{ a }_{ 9 } }$

$=\frac { { \alpha }^{ 10 }-{ \beta }^{ 10 }-2({ \alpha }^{ 8 }-{ \beta }^{ 8 }) }{ { 2(\alpha }^{ 9 }{ -\beta }^{ 9 }) }$

$=\frac { { \alpha }^{ 10 }-{ \beta }^{ 10 }+\alpha \beta ({ \alpha }^{ 8 }-{ \beta }^{ 8 }) }{ { 2(\alpha }^{ 9 }{ -\beta }^{ 9 }) }$

$=\frac { { \alpha }^{ 9 }(\alpha +\beta )-{ \beta }^{ 9 }(\alpha +\beta ) }{ { 2(\alpha }^{ 9 }{ -\beta }^{ 9 }) }$

$=\frac { (\alpha +\beta )({ \alpha }^{ 9 }{ -\beta }^{ 9 }) }{ { 2(\alpha }^{ 9 }{ -\beta }^{ 9 }) }$

$=\frac { (\alpha +\beta ) }{ 2 } =\frac { 6 }{ 2 } =\boxed { 3 }$

$x^2-6x-2=0\quad \Rightarrow x^2=6x+2$

Since $\alpha$ and $\beta$ are roots of $x^2-6x-2=0$

$\Rightarrow \begin{cases} \alpha^2 = 6\alpha + 2 \\ \beta^2 = 6\beta + 2 \end{cases} \Rightarrow \begin{cases} \alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2} \\ \beta^n = 6\beta^{n-1} + 2\beta^{n-2} \end{cases}$

$\Rightarrow \alpha^n - \beta^n = 6(\alpha^{n-1} - \beta^{n-1}) + 2(\alpha^{n-2} - \beta^{n-2})$

$\Rightarrow a_n = 6a_{n-1} + 2a_{n-2}$

Therefore,

$\dfrac {a_{10}}{2a_9} - \dfrac {2a_8}{2a_9} = \dfrac {6a_9 + 2a_8} {2a_9} - \dfrac {a_8}{a_9} = 3 + \dfrac {a_8}{a_9} - \dfrac {a_8}{a_9} = \boxed{3}$