JEE Advance - Mathematics (2011 - Paper 1 Offline - No. 4)
Let $$\alpha $$ and $$\beta $$ be the roots of $${x^2} - 6x - 2 = 0,$$ with $$\alpha > \beta .$$ If $${a_n} = {\alpha ^n} - {\beta ^n}$$ for $$\,n \ge 1$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is
1
2
3
4
Explanation
We have, $${a_n} = {\alpha ^n} - {\beta ^n}$$
$${\alpha ^2} - 6\alpha - 2 = 0$$
Multiplying with $$\alpha$$8 on both sides, we get
$${\alpha ^{10}} - 6{\alpha ^9} - 2{\alpha ^8} = 0$$ ..... (1)
Similarly, $${\beta ^{10}} - 6{\beta ^9} - 2{\beta ^8} = 0$$ ..... (2)
From Eqs. (1) and (2), we get
$${\alpha ^{10}} - {\beta ^{10}} - 6({\alpha ^9} - {\beta ^9}) = 2({\alpha ^8} - {\beta ^8})$$
$$ \Rightarrow {a_{10}} - 6{a_9} = 2{a_8} \Rightarrow {{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$$.
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