Think about Question 1 Let $x_{1}=4$, and $x_{n}=\frac{5 n+2}{5 n-3} x_{n-1}+7(5 n+2)(n=2,3, \cdots)$, find the general term $x_{n}$.
Solution: From the given transformation, we have $\frac{x_{n}}{5 n+2}=\frac{x_{n-1}}{5 n-3} x_{n-1}+7$. Therefore, the sequence $\left\{\frac{x_{n}}{5 n+2}\right\}$ is an arithmetic sequence with the first term $\frac{x_{1}}{5 n+2}=\frac{4}{7}$ and a common difference of 7. Thus, $\frac{x_{n}}{5 n+2}=\frac{4}{7}+7(n-1)$, which means
$$
x_{n}=\frac{1}{7}(49 n-45)(5 n+2) \text {. }
$$