Solve the following
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.

Solution:

Here, $x^{a}=(x z)^{\frac{b}{2}}=z^{c}$

Now, taking log on both the sides:

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

$\Rightarrow \operatorname{alog} \mathrm{x}=\frac{b}{2} \log (\mathrm{xz})=\mathrm{c} \log \mathrm{z}$

$\Rightarrow \operatorname{alog} \mathrm{x}=\frac{b}{2} \log \mathrm{x}+\frac{b}{2} \log \mathrm{z}=\mathrm{c} \log \mathrm{z}$

$\Rightarrow$ alog $\mathrm{x}=\frac{b}{2} \log \mathrm{x}+\frac{b}{2} \log \mathrm{z}$ and $\frac{b}{2} \log \mathrm{x}+\frac{b}{2} \log \mathrm{z}=\mathrm{c} \log \mathrm{z}$

$\Rightarrow\left(\mathrm{a}-\frac{b}{2}\right) \log \mathrm{x}=\frac{b}{2} \log \mathrm{z}$ and $\frac{b}{2} \log \mathrm{x}=\left(\mathrm{c}-\frac{b}{2}\right) \log \mathrm{z}$

$\Rightarrow \frac{\log \mathrm{x}}{\log \mathrm{z}}=\frac{\frac{b}{2}}{\left(\mathrm{a}-\frac{b}{2}\right)}$ and $\frac{\log \mathrm{x}}{\log \mathrm{z}}=\frac{\left(\mathrm{c}-\frac{b}{2}\right)}{\frac{b}{2}}$

$\Rightarrow \frac{\frac{b}{2}}{\left(\mathrm{a}-\frac{b}{2}\right)}=\frac{\left(\mathrm{c}-\frac{b}{2}\right)}{\mathrm{b} / 2}$

$\Rightarrow \frac{\mathrm{b}^{2}}{4}=\mathrm{ac}-\frac{\mathrm{ab}}{2}-\frac{\mathrm{bc}}{2}+\frac{\mathrm{b}^{2}}{4}$

$\Rightarrow 2 \mathrm{ac}=\mathrm{ab}+\mathrm{bc}$

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

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

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