Let - and - be the roots of x2-6x-2 -0 with - If an - -n - -n for n-1- then the value of a10-2a82a9 is

Α^2 6α 2=0 α^2 2=6α β^2 6β 2=0 β^2 2=6β a_10 2a_82a_9=(α^10 β^10) 2(α^8 β^8)2(α^9 β^9) =α^8(α^2 2) β^8(β^2 2)2(α^9 β^9) =α^8(6α ) β^8(6β )2(α^9 β^9) =3 ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

Let α and β b...

Question

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 $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is

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Solution

The correct option is C
$3$

$α^{2} - 6 α - 2 = 0 α^{2} - 2 = 6 α β^{2} - 6 β - 2 = 0 β^{2} - 2 = 6 β \frac{a_{10} - 2 a_{8}}{2 a_{9}} = \frac{(α^{10} - β^{10}) - 2 (α^{8} - β^{8})}{2 (α^{9} - β^{9})} = \frac{α^{8} (α^{2} - 2) - β^{8} (β^{2} - 2)}{2 (α^{9} - β^{9})} = \frac{α^{8} (6 α) - β^{8} (6 β)}{2 (α^{9} - β^{9})} = 3$

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Similar questions

Q. Let

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

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α > β

. lf

a_{n} = α^{n} - β^{n}

for

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\frac{a_{10} - 2 a_{8}}{2 a_{9}}

Q. Let

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and

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be the roots of equation

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a_{n} = α^{n} - β^{n}

, for

n \geq 1

, then the value of

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\frac{a_{10} - 2 a_{8}}{2 a_{9}}

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, for

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Let α and β be the roots of x2−6x−2=0 with α>β. If an=αn−βn for n≥1, then the value of a10−2a82a9 is

The correct option is C 3 α2−6α−2=0α2−2=6αβ2−6β−2=0β2−2=6βa10−2a82a9=(α10−β10)−2(α8−β8)2(α9−β9)=α8(α2−2)−β8(β2−2)2(α9−β9)=α8(6α)−β8(6β)2(α9−β9)=3

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 $\frac{a_{10} - 2 a_{8}}{2 a_{9}}$ is