Let $\alpha$ and $\beta$ be the roots of $x^{2}-6 x-2=0$, with $\alpha>\beta$. If $a_{n}=\alpha^{n}-\beta^{n}$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_{8}}{2 a_{9}}$ is

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Sequences and Series

Let

α

and

β

be the roots of

x^{2} - 6 x - 2 = 0

, with

α > β

. If

a_{n} = α^{n} - β^{n}

for

n \geq 1

, then the value of

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

(A)1
(B)2
(C)3✓ Correct
(D)4

Answer: (C)

Solution

Since

α

and

β

are roots of

x^{2} - 6 x - 2 = 0

, we have

α^{2} = 6 α + 2

and

β^{2} = 6 β + 2

.

Multiplying the first by

α^{8}

α^{10} = 6 α^{9} + 2 α^{8}

.

Similarly:

β^{10} = 6 β^{9} + 2 β^{8}

.

Subtracting:

α^{10} - β^{10} = 6 (α^{9} - β^{9}) + 2 (α^{8} - β^{8})

.

That is,

a_{10} = 6 a_{9} + 2 a_{8}

, so

a_{10} - 2 a_{8} = 6 a_{9}

.

Therefore

\frac{a _{10} - 2 a _{8}}{2 a _{9}} = \frac{6 a _{9}}{2 a _{9}} = 3

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option

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