Question

If $x^a=x^{b/2}{z}^{b/2}=z^c$ then prove that $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P

Answer 1

Given,

$x^a=x^{\frac{b}{2}}{z}^{\frac{b}{2}}=z^c$

$x^a=(xz{)}^{\frac{b}{2}}$ and $z^c=(xz{)}^{\frac{b}{2}}$

$\log x^a=\log (xz{)}^{\frac{b}{2}}$ and $\log z^c=\log (xz{)}^{\frac{b}{2}}$

$a\log x=\frac{b}{2}(\log x+\log z)$ and $c\log z=\frac{b}{2}(\log x+\log z)$

$a=\frac{b}{2}(\log x+\log z)\log x$ and $c=\frac{b}{2}(\log x+\log z)\log z$

$\frac{1}{a}=\frac{1}{\frac{b}{2}(\log x+\log z)\log x}$ and $\frac{1}{c}=\frac{1}{\frac{b}{2}(\log x+\log z)\log x}$

$\frac{1}{a}+\frac{1}{c}=\frac{2\log x}{b(\log x+\log z)}+\frac{2\log z}{b(\log x+\log z)}$

$\frac{1}{a}+\frac{1}{c}=\frac{2(\log x+\log z)}{b(\log x+\log z)}$

$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}=\frac{1}{b}+\frac{1}{b}$

$\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$

$\therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P