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

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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}$

$3b=5c$…………….(i)

Also a,b,c are in GP

So, $b^2=ac$…………..(ii)

From Eqns. (i) and (ii), we get

$9ac=25c^2$

$9a=25c$

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.

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