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Question
If xa = xb/2 zb/2 = zc, then prove that \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P.
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Solution
\[\text { Here }, x^a = \left( xz \right)^\frac{b}{2} = z^c \]
\[\text { Now, taking log on both the sides: } \]
\[\log \left( x \right)^a = \log \left( xz \right)^\frac{b}{2} = \log \left( z \right)^c \]
\[ \Rightarrow \text { alog} x = \frac{b}{2} \log\left( xz \right) = c \log z\]
\[ \Rightarrow \text {alog } x = \frac{b}{2}\log x + \frac{b}{2}\log z = c \log z\]
\[ \Rightarrow \text { alog } x = \frac{b}{2}\log x + \frac{b}{2}\log z \text { and }\frac{b}{2}\log x + \frac{b}{2}\log z = c \log z\]
\[ \Rightarrow \left( a - \frac{b}{2} \right)\log x = \frac{b}{2} \log z \text { and }\frac{b}{2}\log x = \left( c - \frac{b}{2} \right)\log z\]
\[ \Rightarrow \frac{\log x}{\log z} = \frac{\frac{b}{2}}{\left( a - \frac{b}{2} \right)} \text { and } \frac{\log x}{\log z} = \frac{\left( c - \frac{b}{2} \right)}{\frac{b}{2}} \]
\[ \Rightarrow \frac{\frac{b}{2}}{\left( a - \frac{b}{2} \right)} = \frac{\left( c - \frac{b}{2} \right)}{\frac{b}{2}}\]
\[ \Rightarrow \frac{b^2}{4} = ac - \frac{ab}{2} - \frac{bc}{2} + \frac{b^2}{4}\]
\[ \Rightarrow 2ac = ab + bc\]
\[ \Rightarrow \frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]
\[\text { Thus, } \frac{1}{a}, \frac{1}{b} \text { and } \frac{1}{c} \text { are in A . P } .\]
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