Logressions

First lets use the properties of logarithms to simplify the expression:

$\large {\log{(x+z -2y)(x+z)} = \log {(x-z)^2}}$

$\therefore [(x+z)-2y](x+z) =(x-z)^2$

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

$[(x+z)^2 - (x-z)^2] =2y (x+z)$

$\large {2xz= y (x+z)}\huge{ \longrightarrow y=\dfrac {2xz}{ x+z}}$

Now we have to substitute in variables in a and d for each variable , After trial and error we get that the H.P satisfies this. Lets look at the proof.

Now let $\large {x=\frac {1}{a};y=\frac {1}{a+d};z=\frac {1}{a+2d}}$

$\large {\therefore \dfrac{1}{a+d}=\dfrac {2(\frac {1}{a})(\frac {1}{a+2d})}{(\frac {1}{a})+(\frac {1}{a+2d})}}$

$\large { \dfrac{1}{a+d}=\dfrac {2}{a (a+2d)} × \dfrac {1}{\frac {2 (a+d)}{a (a+2d)}}}$

Now simplifying this expression we get:

$\boxed {\dfrac{1}{a+d}=\dfrac{1}{a+d}}$

$\large {\longrightarrow \text {x,y,z are in Harmonic Progression}\longleftarrow}$