Let ${\mathrm{P}}_{\mathrm{n}}={\alpha }^{\mathrm{n}}+{\beta }^{\mathrm{n}},\mathrm{n}\in \mathrm{N}$. If ${\mathrm{P}}_{10}=123,{\mathrm{P}}_{9}=76$, ${\mathrm{P}}_{8}=47$ and ${\mathrm{P}}_{1}=1$, then the quadratic equation having roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is : [JEE Main 2025, 2 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: b) ${x}^{2}+x-1=0$. $S=\alpha+\beta$ $p=\alpha\beta$ $P_1=\alpha+\beta=1$ $S=1$ $P_n=sP_{n-1}-pP_{n-2}$ $P_{10}=123,\quad P_9=76,\quad P_8=47$ $123=P_9-pP_8$ $123=76-p(47)$ $47p=-47$ $p=-1$ $\alpha+\beta=1$ $\alpha\beta=-1$. Therefore $\alpha$ and $\beta$ are roots of $x^2-x-1=0$ Now the roots required are $\dfrac1\alpha$ and $\dfrac1\beta$ Their sum is $\dfrac{\alpha+\beta}{\alpha\beta} =\dfrac{1}{-1}=-1$ Their product is $\dfrac1{\alpha\beta}=-1$ Hence the required quadratic equation is $x^2-(\text{sum})x+(\text{product})=0$ $x^2+x-1=0$

Let \({\mathrm{P}}_{\mathrm{n}}={\alpha }^{\mathrm{n}}+{\beta }^{\mathrm{n}},\mathrm{n}\in \mathrm{N}\). If \({\mathrm{P…

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