If a, b , and c are in A.P. and also a - 2 b, 2 b- 3 c, 3 c- a are...

(\log \mathrm{a}, \log \mathrm{b}, \log \mathrm{c}) are in A.P. ($$\begin{aligned} &\Rightarrow 2 \log \mathrm{b}=\log \mathrm{a}+\log \mathrm{c} \\ &\Rightarrow \log \mathrm{b}^{2}=\log (\mathrm{ac}) \\ &\Rightarrow \mathrm{b}^{2}=\mathrm{ac} \Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c} \text{ are in G.P. } \\ &\log \mathrm{a}-\log 2 \mathrm{~b}, \log 2 \mathrm{~b}-\log 3 \mathrm{c}, \log 3 \mathrm{c}-\log \text{ a are in A.P. } \end{aligned}$$) (\Rightarrow 2(\log 2 b-\log 3 c)=(\log a-\log 2 b)+(\log 3 c-\log a)) (\Rightarrow 3 \log 2 \mathrm{~b}=3 \log 3 \mathrm{c} \Rightarrow 2 \mathrm{~b}=3 \mathrm{c}) Now, (b^{2}=a c \Rightarrow b^{2}=a \cdot \frac{2 b}{3} \Rightarrow b=\frac{2 a}{3}, c=\frac{4 a}{9}) ($$\begin{aligned} &\text{ i.e., } a=a, b=\frac{2 a}{3}, c=\frac{4 a}{9} \\ &\Rightarrow a: b: c=1: \frac{2}{3}: \frac{4}{9}=9: 6: 4 \end{aligned}$$) Since, sum of any two is greater than the (3^{\mathrm{rd}}, \mathrm{a}, \mathrm{b}, \mathrm{c}) form a triangle.