If x, 2y, 3z are in A.P., where the distinct numbers x, y, z are in G.P.,

Solution:

Let x, 2y, 3z be in A.P, for distinct x, y and z

where x, y, z are in G.P

Since x, 2y and 3z are in A.P

$\therefore 2 y=\frac{x+3 y}{2} \quad(\because$ middle term $=$ arithmetic mean of adjacent terms)

i. e $4 y=x+3 y \quad \ldots(1)$

and since x, y and z are in G.P

∴ y2 = xz       ...(2)

Let r denote the common ratio of G.P

$\therefore \frac{y}{x}=r$ and $\frac{z}{x}=r^{2} \quad \ldots(3)$

Now, dividing (1) by x, we get

$\frac{4 y}{x}=1+\frac{3 y}{x}$

i.e $4 r=1+3 r^{2}$

i. e $3 r^{2}-4 r+1=0$ i. e $3 r^{2}-3 r-r+1=0$

i. e $(3 r-1)(r-1)=0$

i.e $r=\frac{1}{3}$ or $r=1$

(r = 1 is not possible since x, y, z are distinct)

$\Rightarrow$ Common ratio is $\frac{1}{3}$

Hence, the correct answer is option B.