If a, b, c are in G.P. and log a - log 2b, lo | vedclass

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(B) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A.P.$,the sum of the f

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Obtuse angled

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(B) Given that $a, b, c$ are in $G.P.$,we have $b^2 = ac$.Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A.P.$,the sum of the first and third terms equals twice the middle term:$(\log a - \log 2b) + (\log 3c - \log a) = 2(\log 2b - \log 3c)$$\log 3c - \log 2b = 2\log 2b - 2\log 3c$$3\log 3c = 3\log 2b \Rightarrow 3c = 2b \Rightarrow b = \frac{3}{2}c$.Substituting $b = \frac{3}{2}c$ into $b^2 = ac$,we get $(\frac{3}{2}c)^2 = ac \Rightarrow \frac{9}{4}c^2 = ac \Rightarrow a = \frac{9}{4}c$.Thus,the sides are $a = \frac{9}{4}c, b = \frac{3}{2}c, c = c$.Multiplying by $\frac{4}{c}$,the sides are proportional to $9, 6, 4$.Since $9^2 > 6^2 + 4^2$ $(81 > 36 + 16 = 52)$,the triangle is obtuse-angled.

If $a, b, c$ are in $G.P.$ and $\log a - \log 2b, \log 2b - \log 3c$ and $\log 3c - \log a$ are in $A.P.$,then $a, b, c$ are the lengths of the sides of a triangle which is

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