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Consider the statement $ \log \bigg(\large\frac{5c}{a} \bigg),\: \log \bigg(\large\frac{3b}{5c} \bigg)$ and $ \log \bigg(\large\frac{a}{3b} \bigg)$ are in AP, where a,b,c are in GP. The statement is true only if a,b,c are the lengths of sides of

$\begin {array} {1 1} (A)\;an\: isosceles \: triangle & \quad (B)\;an\: equilateral\: triangle \\ (C)\;a\: scalene\: triangle & \quad (D)\;None \: of\: these \end {array}$

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Ans : (D)
Since, $ \log \bigg( \large\frac{5c}{a} \bigg), \: \log \bigg( \large\frac{3b}{5c} \bigg) \: and \: \log \bigg( \large\frac{a}{3b} \bigg)$ are in AP,
So, $2\log \bigg( \large\frac{3b}{5c} \bigg) = \log \bigg( \large\frac{5c}{a} \bigg) + \log \bigg( \large\frac{a}{3b} \bigg)$
$ \bigg( \large\frac{3b}{5c} \bigg)^2=\large\frac{5c}{a} \: \large\frac{ a}{3b}$
Also a,b,c are in GP
So, $b^2=ac$…………..(ii)
From Eqns. (i) and (ii), we get
So, $9.a/5=5c=3b$ [From Eq. (i)]
$ \large\frac{a}{5}=\large\frac{b}{3}=\large\frac{c}{ \bigg( \large\frac{9}{5} \bigg)}$
$b+c < a$
Since, sum of two sides of a triangle is smaller than third side,
Therefore, triangle is not defined.
answered Jan 23, 2014 by thanvigandhi_1

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