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

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(B) Given $a, b, c$ are in $G$.$P$.,so $b^2 = ac$. Let $r = \frac{b}{a} = \frac{c}{b}$, then $b = ar$ and $c = ar^2$. Since $\log a - \log 2b, \log 2b

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obtuse angled

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(B) Given $a, b, c$ are in $G$.$P$.,so $b^2 = ac$. Let $r = \frac{b}{a} = \frac{c}{b}$, then $b = ar$ and $c = ar^2$. Since $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A$.$P$.,we have $2(\log 2b - \log 3c) = (\log a - \log 2b) + (\log 3c - \log a)$. $2 \log(\frac{2b}{3c}) = \log(\frac{a}{2b} 	imes \frac{3c}{a}) = \log(\frac{3c}{2b})$. Let $x = \frac{2b}{3c}$, then $2 \log x = \log(\frac{1}{x}) = -\log x$, which implies $3 \log x = 0$, so $x = 1$. Thus, $\frac{2b}{3c} = 1 \Rightarrow 2b = 3c$. Since $b = ar$ and $c = ar^2$, we have $2(ar) = 3(ar^2)$ $\Rightarrow 2 = 3r$ $\Rightarrow r = \frac{2}{3}$. The sides are $a, b = \frac{2a}{3}, c = \frac{4a}{9}$. For a triangle, the sum of two smaller sides must be greater than the third side: $\frac{2a}{3} + \frac{4a}{9} = \frac{10a}{9} > a$. This holds. Using the Cosine rule for the largest angle $A$ (opposite to side $a$): $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(\frac{2a}{3})^2 + (\frac{4a}{9})^2 - a^2}{2(\frac{2a}{3})(\frac{4a}{9})} = \frac{\frac{4}{9} + \frac{16}{81} - 1}{\frac{16}{27}} = \frac{\frac{36+16-81}{81}}{\frac{16}{27}} = \frac{-29}{81} 	imes \frac{27}{16} = -\frac{29}{48} Since $\cos A < 0$, the angle $A$ is obtuse.

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

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