lf a,b,c  are  in GP and $\log \mathrm{a}-\log 2\mathrm{b},\log 2\mathrm{b}-\log 3\mathrm{c}$,and $\log 3\mathrm{c}-\log \mathrm{a}$are in AP,  then  o,b,c  are the length  of the  sides  of a triangle  which is

Since, a ,b ,c  are in GP. 

$\Rightarrow {\mathrm{b}}^2=\mathrm{ac}$ 

and log$\mathrm{\alpha }$ -log2 b,log2b-log3c and log3c-log$\mathrm{\alpha }$ are in AP. 

$\begin{array}{r}\Rightarrow 2(\log 2\mathrm{b}-\log 3\mathrm{c})=\log \mathrm{a}-\log 2\mathrm{b}+\log 3\mathrm{c}-\log \mathrm{a}\\ \therefore {\mathrm{b}}^2=\mathrm{ac}\text{ and }2\mathrm{b}=3\mathrm{c}\\ \Rightarrow \mathrm{b}=\frac{2\mathrm{a}}{3}\text{ and }\mathrm{c}=\frac{4\mathrm{a}}{9}\\ \text{ Since, }\mathrm{a}+\mathrm{b}=\frac{5\mathrm{a}}{3}>\mathrm{c},\mathrm{b}+\mathrm{c}=\frac{10\mathrm{a}}{9}>\mathrm{a}\\ \mathrm{c}+\mathrm{a}=\frac{13\mathrm{a}}{9}>\mathrm{b}\end{array}$