by newton's theorem $\begin{aligned}
& \mathrm{a}{\mathrm{n}+2}-\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}{\mathrm{n}+1}+\mathrm{a}{\mathrm{n}}=0 \
& \therefore \mathrm{a}{2025}+\mathrm{a}{2023}=\left(\mathrm{t}^2-5 \mathrm{t}+6\right) \mathrm{a}{2024} \
& \therefore \frac{\mathrm{a}{2025}+\mathrm{a}{2023}}{\mathrm{a}_{2024}}=\mathrm{t}^2-5 \mathrm{t}+6 \
& \because \mathrm{t}^2-5 \mathrm{t}+6=\left(\mathrm{t}-\frac{5}{2}\right)^2-\frac{1}{4} \
& \therefore \text { minimum value }=-\frac{1}{4}
\end{aligned}$