SATHEE: Theory of Equations 1 Question 20

Theory of Equations 1 Question 20

21. 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

(a) 1

(b) 2

(d) 4

(2011)

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Answer:

Correct Answer: 21. (c)

Solution:

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

$= \frac{α^{8} (α^{2} - 2) - β^{8} (β^{2} - 2)}{2 (α^{9} - β^{9})}$

$∵ α$ is root of $x^{2} - 6 x - 2 = 0 \Rightarrow α^{2} - 2 = 6 α$

[and $β$ is root of $x^{2} - 6 x - 2 = 0 \Rightarrow β^{2} - 2 = 6 β$ ]

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

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Theory of Equations 1 Question 20

21. 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

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21. 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