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If $a, b, c$ are in G. P. and $\log a .-\log 2 b, \log 2 b-\log 3 c, \log 3 c-\log a$ are in A. P., then $a, b, c$ are the lengths of the sides of a triangle which is
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The correct answer is:
obtuse angled
$$
\because 2 \log \frac{2 b}{3 c}=\log \frac{a}{2 b}+\log \frac{3 c}{a} \Rightarrow \frac{4 b^2}{9 c^2}=\frac{3 c}{2 b} \Rightarrow 2 b=3 c \Rightarrow \frac{c}{b}=\frac{2}{3} \text { (common ratio) }
$$
$\therefore$ Sides are a, $\frac{2 \mathrm{a}}{3}, \frac{4 \mathrm{a}}{9}$
Using Cosine rule
$\Rightarrow \cos \mathrm{A} < 0 \Rightarrow$ obteuse angled tiangle
\because 2 \log \frac{2 b}{3 c}=\log \frac{a}{2 b}+\log \frac{3 c}{a} \Rightarrow \frac{4 b^2}{9 c^2}=\frac{3 c}{2 b} \Rightarrow 2 b=3 c \Rightarrow \frac{c}{b}=\frac{2}{3} \text { (common ratio) }
$$
$\therefore$ Sides are a, $\frac{2 \mathrm{a}}{3}, \frac{4 \mathrm{a}}{9}$
Using Cosine rule
$\Rightarrow \cos \mathrm{A} < 0 \Rightarrow$ obteuse angled tiangle
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