Mathematics - Sequence & Series Question with Solution | TestHub

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