If ${\log }_{e}a,{\log }_{e}b,{\log }_{e}c$ are in an $A.P.$ and ${\log }_{e}a-{\log }_{e}2b,{\log }_{e}2b-{\log }_{e}3c,{\log }_{e}3c-{\log }_{e}a$ are also in an $A.P.$, then $a:b:c$ is equal to

Given: $\log a, \log b, \log c$ are in $A.P.$

$\Rightarrow 2\log b=\log a+\log c$

$\Rightarrow b^2=ac ...\left(1\right)$

And $\log a-\log 2b, \log 2b-\log 3c, \log 3c-\log a$ are in $A.P$ 

$\Rightarrow 2\log \frac{2b}{3c}=\log \frac{a}{2b}+\log \frac{3c}{a}$

$\Rightarrow {\left(\frac{2b}{3c}\right)}^2=\frac{a}{2b}\times \frac{3c}{a}$

$\Rightarrow {\left(\frac{2b}{3c}\right)}^2=\frac{3c}{2b}$

$\Rightarrow 2b=3c$ 

$\Rightarrow 4b^2=9c^2 \& 6b=9c ...\left(2\right)$

Now solving the equation $\left(1\right) \& \left(2\right)$ we get,

$4ac=9c^2$

$\Rightarrow 4a=9c ...\left(3\right)$

Then from equation $\left(2\right) \& \left(3\right)$ we get,

$4a=6b=9c=k$

$\Rightarrow a=\frac{k}{4}, b=\frac{k}{6} \& c=\frac{k}{9}$

$\Rightarrow a:b:c=\frac{1}{4}:\frac{1}{6}:\frac{1}{9}=9:6:4$