JEE Main 2025 (Online) 2nd April Morning ShiftQuestion 1
Let $P_n=\alpha^{n}+\beta^{n}$, $n\in\mathbf{N}$. If $P_{10}=123$, $P_9=76$, $P_8=47$ and $P_1=1$, then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is :
Quadratic Equation · Powers of Roots (αⁿ ± βⁿ Recurrence)
Solution
Concept. $P_{10}$, $P_9$, $P_8$ are linked by the recurrence; that gives $\alpha\beta$. Formulas used. (1) $P_{n+2}=(\alpha+\beta)P_{n+1}-\alpha\beta\,P_n$(2) Equation with roots $\frac1\alpha,\frac1\beta$: $x^2-\dfrac{\alpha+\beta}{\alpha\beta}x+\dfrac{1}{\alpha\beta}=0$Step 1 - Find $\alpha\beta$. $\alpha+\beta=P_1=1$. By (1), $123=76-47\,\alpha\beta$, so $\alpha\beta=-1$. Step 2 - New equation. By (2): sum $=\frac{1}{-1}=-1$, product $=-1$: $$x^2+x-1=0.$$ Answer: (B)