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

$$\begin{gathered} \mathrm{Here},x^a\mathit{=}\mathit{}{\left(xz\right)}^{\frac{b}{2}}\mathit{}\mathit{=}\mathit{}{z}^c \\ \mathrm{Now},\mathrm{taking}\log \mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ \log {\left(\mathrm{x}\right)}^{\mathrm{a}}=\log {\left(\mathrm{xz}\right)}^{\frac{b}{2}}=\log {\left(\mathrm{z}\right)}^{\mathrm{c}} \\ \Rightarrow \mathrm{alog}\mathrm{x}=\frac{b}{2}\log \left(\mathrm{xz}\right)=\mathrm{c}\log \mathrm{z} \\ \Rightarrow \mathrm{alog}\mathrm{x}=\frac{b}{2}\log \mathrm{x}+\frac{b}{2}\log \mathrm{z}=\mathrm{c}\log \mathrm{z} \\ \Rightarrow \mathrm{alog}\mathrm{x}=\frac{b}{2}\log \mathrm{x}+\frac{b}{2}\log \mathrm{z}\mathrm{and}\frac{b}{2}\log \mathrm{x}+\frac{b}{2}\log \mathrm{z}=\mathrm{c}\log \mathrm{z} \\ \Rightarrow \left(\mathrm{a}-\frac{b}{2}\right)\log \mathrm{x}=\frac{b}{2}\log \mathrm{z}\mathrm{and}\frac{b}{2}\log \mathrm{x}=\left(\mathrm{c}-\frac{b}{2}\right)\log \mathrm{z} \\ \Rightarrow \frac{\log \mathrm{x}}{\log \mathrm{z}}=\frac{{\displaystyle \frac{b}{2}}}{\left(\mathrm{a}-{\displaystyle \frac{b}{2}}\right)}\mathrm{and}\frac{\log \mathrm{x}}{\log \mathrm{z}}=\frac{\left(\mathrm{c}-{\displaystyle \frac{b}{2}}\right)}{{\displaystyle \frac{b}{2}}} \\ \Rightarrow \frac{{\displaystyle \frac{b}{2}}}{\left(\mathrm{a}-{\displaystyle \frac{b}{2}}\right)}=\frac{\left(\mathrm{c}-{\displaystyle \frac{b}{2}}\right)}{{\displaystyle \raisebox{1ex}{$\mathrm{b}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \\ \Rightarrow \frac{{\mathrm{b}}^2}{4}=\mathrm{ac}-\frac{\mathrm{ab}}{2}-\frac{\mathrm{bc}}{2}+\frac{{\mathrm{b}}^2}{4} \\ \Rightarrow 2\mathrm{ac}=\mathrm{ab}+\mathrm{bc} \\ \Rightarrow \frac{2}{\mathrm{b}}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{c}} \\ \mathrm{Thus},\frac{1}{\mathrm{a}},\frac{1}{\mathrm{b}}\mathrm{and}\frac{1}{\mathrm{c}}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \end{gathered}$$