Area Stage

Solution:
Let $3x$, $4y$, $5z$ be in geometric progression. Then there exists a common ratio $r$ such that:

$$4y = 3x \cdot r, \quad 5z = 4y \cdot r$$

From the first equation:
$$4y = 3x r \implies y = \frac{3x r}{4}$$
From the second equation:
$$5z = 4y r \implies z = \frac{4y r}{5}$$
Substitute $y$ from above:
$$z = \frac{4}{5} \cdot \frac{3x r}{4} \cdot r = \frac{3x r^2}{5}$$

Now, $\frac{1}{x}$, $\frac{1}{y}$, $\frac{1}{z}$ are in arithmetic progression. So:
$$2 \cdot \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$$
Substitute $y$ and $z$ in terms of $x$ and $r$:
$$2 \cdot \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$$
$$2 \cdot \frac{1}{\frac{3x r}{4}} = \frac{1}{x} + \frac{1}{\frac{3x r^2}{5}}$$
$$2 \cdot \frac{4}{3x r} = \frac{1}{x} + \frac{5}{3x r^2}$$
$$\frac{8}{3x r} = \frac{1}{x} + \frac{5}{3x r^2}$$
Multiply both sides by $3x r^2$:
$$8 r = 3 r^2 + 5$$
$$3 r^2 - 8 r + 5 = 0$$
Solve this quadratic equation for $r$:
$$r = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm 2}{6}$$
So $r = \frac{10}{6} = \frac{5}{3}$ or $r = \frac{6}{6} = 1$

If $r = 1$, then $3x = 4y = 5z$, which would make $x$, $y$, $z$ all proportional, but then $\frac{1}{x}$, $\frac{1}{y}$, $\frac{1}{z}$ would also be in geometric progression, not necessarily arithmetic unless all are equal. Let's check $r = \frac{5}{3}$.

So $r = \frac{5}{3}$.

Now, recall:
$$y = \frac{3x r}{4} = \frac{3x \cdot \frac{5}{3}}{4} = \frac{5x}{4}$$
$$z = \frac{3x r^2}{5} = \frac{3x \cdot \left(\frac{5}{3}\right)^2}{5} = \frac{3x \cdot \frac{25}{9}}{5} = \frac{25x}{15} = \frac{5x}{3}$$

Now, $\frac{x}{z} + \frac{z}{x}$:
$$\frac{x}{z} + \frac{z}{x} = \frac{x}{\frac{5x}{3}} + \frac{\frac{5x}{3}}{x} = \frac{3}{5} + \frac{5}{3} = \frac{9 + 25}{15} = \frac{34}{15}$$

Therefore, the answer is $\boxed{\dfrac{34}{15}}$.