JEE Main 2025 — Mathematics Algebra

$$\begin{aligned} & x^2+\sqrt{3} \mathrm{x}-16=0 \lt \beta \\ & \mathrm{P}_{\mathrm{n}}+\sqrt{3} \mathrm{P}_{\mathrm{n}-1}-16 \mathrm{P}_{\mathrm{n}-2}=0 \\ & \mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}-16 \mathrm{P}_{23}=0 \\ & \therefore \frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}=8\end{aligned}$$​x2+3​x−16=0<βPn​+3​Pn−1​−16Pn−2​=0P25​+3​P24​−16P23​=0∴2P23​P25​+3​P24​​=8​

Similarly

$$x^2+3 x-1=0 \lt \sum_\delta^\gamma \quad Q_n=\gamma^n+\delta^n$$x2+3x−1=0<δ∑γ​Qn​=γn+δn

$\begin{aligned}
& \mathrm{Q}{25}-\mathrm{Q}{23}=\gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23} \
&=\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \
&=\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \gamma) \
&=-3\left[\gamma^{24}+\delta^{24}\right] \
&=-3 \mathrm{Q}{24} \
& \therefore \frac{\mathrm{Q}{25}-\mathrm{Q}{23}}{\mathrm{Q}{24}}=-3
\end{aligned}$

$$\frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=8-3=5$$2P23​P25​+3​P24​​+Q24​Q25​−Q23​​=8−3=5