If $x^a = y^b = z^c$ and $y^2 = xz$; prove that $b = \dfrac{2ac}{a+c}$.

Let $x^a = y^b = z^c = k$

⇒ $x^a = k$, $y^b = k$, $z^c = k$

⇒ $x = k^{\dfrac{1}{a}}$, $y = k^{\dfrac{1}{b}}$, $z = k^{\dfrac{1}{c}}$

Substitute the values of x, y and z in $y^2 = xz$,

$⇒ \Big(k^{\dfrac{1}{b}}\Big)^2 = k^{\dfrac{1}{a}} \times k^{\dfrac{1}{c}}\\[1em] ⇒ k^{\dfrac{2}{b}} = k^{\dfrac{1}{a} + \dfrac{1}{c}}\\[1em] ⇒ \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\\[1em] ⇒ \dfrac{2}{b} = \dfrac{c + a}{ac} \\[1em] ⇒ 2 \times ac = b(a + c)\\[1em] ⇒ b = \dfrac{2ac}{a + c}$

Hence, $b = \dfrac{2ac}{a+c}$.