Suppose $a, b, c$ are distinct positive real numbers such that $a, 2b, 3c$ are in A.P. and $a, b, c$ are in G.P. The common ratio of G.P. is

We have $2\left(2b\right)=a+3c$ and $b^2=ac$
$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\therefore \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{16}(a+3c{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=ac\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2+9c^2+6ac\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=16ac\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-10ac+9c^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} (a-c)(a-9c)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=0\Rightarrow a=9c[\because a\ne c]\end{array}$
$\Rightarrow \frac{c}{a}=\frac{1}{9}$ or $r^2=\frac{1}{9}\Rightarrow r=\frac{1}{3}$

                                           $[\because a,b,c>0]$