• question_answer 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 length of the sides of a triangle which is A) Acute angled B) Obtuse angled C) Right angled D) Equilateral

As given ${{b}^{2}}=ac$ and $2(\log 2b-\log 3c)=\log a-\log 2b+\log 3c-\log a$ $\Rightarrow$${{b}^{2}}=ac$  and  $2b=3c$$\Rightarrow$$b=2a/3$and$c=4a/9$  Since $a+b=\frac{5a}{3}>c,\ b+c=\frac{10a}{9},>a,\ c+a=\frac{13a}{9}>b$ It implies that $a,\ b,\ c$ form $a$ triangle with $a$ as the greatest side. Now, let us find the greatest angle $A$ of $\Delta ABC$ by using the cosine formula. $\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-\frac{29}{48}<0$ $\therefore$ The angle $A$ is obtuse.