2020/002) a,b,c are in AP, x,y,z are in HP and ax,by,cz are in GP then prove that $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$

We have $x,y,z$ are in HP so
$\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$
Squaring both sides
$\frac{1}{x^2} + \frac{1}{z^2} + \frac{2}{xz} = \frac{4}{y^2}$
Or $\frac{1}{x^2} + \frac{1}{z^2} = \frac{4}{y^2} - \frac{2}{xz}  $
Or $\frac{x^2+z^2}{x^2z^2} = \frac{4}{y^2} - \frac{2}{xz}  $
Or $\frac{x^2+z^2}{xz} = \frac{4xz}{y^2} - 2$\
Or $\frac{x}{z} + \frac{z}{x} = \frac{4xz}{y^2} - 2\cdots(1)$
As ax,by,cz are in GP
$axcz = b^2y^2$
Or $\frac{xz}{y^2} = \frac{b^2}{ac}$
Putting above in (1)
$\frac{x}{z} + \frac{z}{x} = \frac{4b^2}{ac} - 2\cdots(2)$
As a,b,c are in AP we have $2b= a + c$
Putting in (2) we get
$\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2}{ac} - 2$
Or $\frac{x}{z} + \frac{z}{x} = \frac{(a+c)^2-2 ac} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a^2+c^2} {ac}$
Or $\frac{x}{z} + \frac{z}{x} = \frac{a}{c} + \frac{c}{a}$
proved