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

Answer

$\text{x}^{\text{a}}=\text{x}^{\frac{\text{b}}{2}}\text{z}^{\frac{\text{b}}{2}}=\text{z}^{\text{c}}=\lambda(\text{ say})$ $\text{x}=\lambda^{\frac{1}{\text{a}}},\text{z}=\lambda^{\frac{1}{\text{c}}}$ $\text{x}^{\frac{\text{b}}{2}}\times\text{z}^{\frac{\text{b}}{2}}=\lambda$ $\lambda^{\frac{1}{\text{a}}\big(\frac{\text{b}}{2}\big)}\times\lambda^{\frac{\text{b}}{2}\times\frac{1}{\text{c}}}=\lambda$ $\lambda^{\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}}=\lambda^1$ $\frac{\text{b}}{2\text{a}}+\frac{\text{b}}{2\text{c}}=1$ $\frac{1}{\text{a}}+\frac{1}{\text{c}}=\frac{2}{\text{b}}$ $\Rightarrow\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}\text{ are in A.P.}$

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