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Three unequal positive numbers a, b, c are such that a, b, c are in G.P. while $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in A.P. Then $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are the lengths of the sides of
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a scalene triangle
$\log \frac{5 c}{2 a}+\log \frac{2 a}{7 b}=2 \log \frac{7 b}{5 c} \Rightarrow \log \frac{5 c}{7 b}=\log \frac{49 b^{2}}{25 c^{2}}$
$\begin{array}{l}
\Rightarrow 5^{3} \mathrm{c}^{3}=7 \mathrm{~b}^{3} \Rightarrow 5 \mathrm{c}=7 \mathrm{~b} \Rightarrow \mathrm{c}=\frac{7}{5} \mathrm{~b} \\
\because \mathrm{b}^{2}=\mathrm{ac}=\mathrm{a} \cdot \frac{7}{5} \mathrm{~b} \Rightarrow \mathrm{a}=\frac{5 \mathrm{~b}}{7}
\end{array}$
Sides are $\frac{5 b}{7}, b, \frac{7}{5} b$
$\begin{array}{l}
\Rightarrow 5^{3} \mathrm{c}^{3}=7 \mathrm{~b}^{3} \Rightarrow 5 \mathrm{c}=7 \mathrm{~b} \Rightarrow \mathrm{c}=\frac{7}{5} \mathrm{~b} \\
\because \mathrm{b}^{2}=\mathrm{ac}=\mathrm{a} \cdot \frac{7}{5} \mathrm{~b} \Rightarrow \mathrm{a}=\frac{5 \mathrm{~b}}{7}
\end{array}$
Sides are $\frac{5 b}{7}, b, \frac{7}{5} b$
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