If a, b, c are in geometric progression and a, 2b ,3c are in arithmetic progression, then what is the common ratio r such that $0<\mathrm{r}<1?$

Given that a, b, c, are in GP Let r be common ratio of GP. So, $a = a, b = ar$ and $c = a{r}^2$

 Also, given that a, 2b, 3c are in AP. 

$\begin{array}{l}\Rightarrow 2b=\frac{a+3c}{2}\\ \Rightarrow 4b=a+3c .....\left(1\right)\end{array}$

From Eq. (1)

$4ar=a+3a{r}^2$

$\begin{array}{l}\Rightarrow 3r^2-4r+1=0\\ \Rightarrow 3r^2-3r-r+1=0\\ \Rightarrow 3r(r-1)-1(r-1)=0\\ \Rightarrow (r-1)(3r-1)=0\\ \Rightarrow r=1\text{ or }r=\frac{1}{3}\end{array}$