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 $$n \geq 1$$, then the value of $$\displaystyle \frac{a_{10}-2a_{8}}{2a_{9}}$$ is

A
1
B
2
C
3
D
4

Solution

## The correct option is A $$3$$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 $$\displaystyle \frac{a_{10} - 2a_8}{2a_{9}}$$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 $$a_{10} = 6\alpha^{9} + 2\alpha^{8} - (6\beta^{9} + 2\beta^{8})$$$$a_{10} = 6\alpha^{9} + 2\alpha^{8} - 6\beta^{9} - 2\beta^{8}$$      $$= 6(\alpha^{9} - \beta^{9}) + 2(\alpha^{8} - \beta^{8})$$      $$a_{10} = 6a_{9} + 2a_{8}$$      $$a_{10} - 2a_{8} = 6a_{9}$$      $$\displaystyle \frac{a_{10} - 2a_{8}} {2a_{9}} = 3$$Maths

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