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

\[\text { Here }, x^a = \left( xz \right)^\frac{b}{2} = z^c \]

\[\text { Now, taking log on both the sides: } \]

\[\log \left( x \right)^a = \log \left( xz \right)^\frac{b}{2} = \log \left( z \right)^c \]

\[ \Rightarrow \text { alog} x = \frac{b}{2} \log\left( xz \right) = c \log z\]

\[ \Rightarrow \text {alog } x = \frac{b}{2}\log x + \frac{b}{2}\log z = c \log z\]

\[ \Rightarrow \text { alog } x = \frac{b}{2}\log x + \frac{b}{2}\log z \text { and }\frac{b}{2}\log x + \frac{b}{2}\log z = c \log z\]

\[ \Rightarrow \left( a - \frac{b}{2} \right)\log x = \frac{b}{2} \log z \text { and }\frac{b}{2}\log x = \left( c - \frac{b}{2} \right)\log z\]

\[ \Rightarrow \frac{\log x}{\log z} = \frac{\frac{b}{2}}{\left( a - \frac{b}{2} \right)} \text { and } \frac{\log x}{\log z} = \frac{\left( c - \frac{b}{2} \right)}{\frac{b}{2}} \]

\[ \Rightarrow \frac{\frac{b}{2}}{\left( a - \frac{b}{2} \right)} = \frac{\left( c - \frac{b}{2} \right)}{\frac{b}{2}}\]

\[ \Rightarrow \frac{b^2}{4} = ac - \frac{ab}{2} - \frac{bc}{2} + \frac{b^2}{4}\]

\[ \Rightarrow 2ac = ab + bc\]

\[ \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]

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