CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
2
loader
C
3
loader
D
4
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image