IIT-JEE 2011 · Paper 1 · Mathematics Q52

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MathematicsSingle correct+3 / -1Question 52 of 69

Let $\alpha$ and $\beta$ be the roots of $x^2-6x-2=0$ , with $\alpha>\beta$ . If $a_n=\alpha^n-\beta^n$ for $n\ge1$ , then the value of $\dfrac{a_{10}-2a_8}{2a_9}$ is

Answer and worked solution

Official answer: C

Recurrence. $\alpha^2=6\alpha+2$ and $\beta^2=6\beta+2$ , so $\alpha^{10}=6\alpha^9+2\alpha^8$ (likewise for β). Subtracting, $a_{10}=6a_9+2a_8$ .

Value. $\dfrac{a_{10}-2a_8}{2a_9}=\dfrac{6a_9}{2a_9}=3$ .

(C)

Topics:Roots and coefficients Special series sums

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IIT-JEE 2011 · Paper 1, Mathematics question 52