Given that if $x, y, z$ form a harmonic progression, then $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ should form an arithmetic progression, i.e.,

$$\frac{2}{y}=\frac{1}{x}+\frac{1}{z}$$

or

$$2 x z=y(x+z)$$

Consider the expression:

$$\begin{gathered} \lg (x+z)+\lg (x-2 y+z)=\lg (x+z)(x-2 y+z)= \\ =\lg \left[(x+z)^{2}-2 y(x+z)\right]=\lg \left[(x+z)^{2}-4 x z\right]= \\ =\lg (x-z)^{2}=2 \lg (x-z) \end{gathered}$$

since, taking into account equation (1), we have the right to write $4 x z$ instead of $2 y(x+z)$.