Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3 x}-16=0$, and $\gamma$ and $\delta$ be the roots of…
- $3$
- $4$
- $5$
- $7$
Solution
Similarly
$x^2+3 x-1=0 \lt \sum_\delta^\gamma \quad Q_n=\gamma^n+\delta^n$
$\begin{aligned}
& \mathrm{Q}_{25}-\mathrm{Q}_{23}=\gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23} \\ &=\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\ &=\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \gamma) \\ &=-3\left[\gamma^{24}+\delta^{24}\right] \\ &=-3 \mathrm{Q}_{24} \\ & \therefore \frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=-3
\end{aligned}$
$\frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=8-3=5$ /
Asked in: JEE Main 2025 (03 Apr Shift 1)