7. Let $x, y, z$ be real numbers, $3 x, 4 y, 5 z$ form a geometric sequence, and $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ form an arithmetic sequence, then the value of $\frac{x}{z}+\frac{z}{x}$ is $\qquad$
7. $\frac{34}{15}$

From the problem, we have $\left\{\begin{array}{l}\frac{2}{y}=\frac{1}{x}+\frac{1}{z} \\ (4 y)^{2}=15 x z\end{array}\right.$, eliminating $y$ gives $16\left(\frac{2 x z}{x+z}\right)^{2}=15 x z$, which simplifies to $\frac{(x+z)^{2}}{x z}=\frac{64}{15}$, thus $\frac{x}{z}+\frac{z}{x}=\frac{34}{15}$.