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(C) Since $\log \left(\frac{5 c}{2 a}\right), \log \left(\frac{7 b}{5 c}\right), \log \left(\frac{2 a}{7 b}\right)$ are in $A.P.$,we have: $\log \left

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a scalene triangle

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(C) Since $\log \left(\frac{5 c}{2 a}ight), \log \left(\frac{7 b}{5 c}ight), \log \left(\frac{2 a}{7 b}ight)$ are in $A.P.$,we have: $\log \left(\frac{5 c}{2 a}ight) + \log \left(\frac{2 a}{7 b}ight) = 2 \log \left(\frac{7 b}{5 c}ight)$ $\log \left(\frac{5 c}{2 a} 	imes \frac{2 a}{7 b}ight) = \log \left(\frac{7 b}{5 c}ight)^2$ $\log \left(\frac{5 c}{7 b}ight) = \log \left(\frac{49 b^2}{25 c^2}ight)$ $\frac{5 c}{7 b} = \frac{49 b^2}{25 c^2}$ $\Rightarrow 125 c^3 = 343 b^3$ $\Rightarrow 5 c = 7 b$ $\Rightarrow c = \frac{7}{5} b$ Since $a, b, c$ are in $G.P.$,$b^2 = ac$. Substituting $c = \frac{7}{5} b$,we get $b^2 = a \left(\frac{7}{5} bight) \Rightarrow a = \frac{5}{7} b$. The sides are $\frac{5}{7} b, b, \frac{7}{5} b$. Let $b = 35k$. Then the sides are $25k, 35k, 49k$. Since $25k + 35k = 60k > 49k$,$25k + 49k = 74k > 35k$,and $35k + 49k = 84k > 25k$,these sides form a triangle. Since all sides are unequal,it is a scalene triangle.

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 $a, b, c$ are the lengths of the sides of

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