jmc

Since $a,$ $b,$ $c$ form a geometric sequence, $b=\sqrt{ac}.$ Then the three logarithms become $${\log }_{c}a,{\log }_{\sqrt{ac}}c,{\log }_a\sqrt{ac}.$$Let $x={\log }_{c}a.$ Then by the change-of-base formula, $${\log }_{\sqrt{ac}}c=\frac{{\log }_{c}c}{{\log }_c\sqrt{ac}}=\frac{1}{\frac{1}{2}{\log }_{c}ac}=\frac{2}{{\log }_{c}a+{\log }_{c}c}=\frac{2}{x+1},$$and $${\log }_a\sqrt{ac}=\frac{1}{2}{\log }_{a}ac=\frac{{\log }_{c}ac}{2{\log }_{c}a}=\frac{{\log }_{c}a+{\log }_{c}c}{2{\log }_{c}a}=\frac{x+1}{2x}.$$Let $d$ be the common difference, so $$d=\frac{2}{x+1}-x=\frac{x+1}{2x}-\frac{2}{x+1}.$$Then $$4x-2x^2(x+1)=(x+1{)}^2-4x,$$which simplifies to $2x^3+3x^2-6x+1=0.$ This factors as $(x-1)(2x^2+5x-1)=0.$

If $x=1,$ then ${\log }_{c}a=1,$ so $a=c.$ But $a$ and $c$ are distinct, so $2x^2+5x-1=0,$ so $x^2=\frac{1-5x}{2}.$ Then $$d=\frac{2}{x+1}-x=\frac{2-x^2-x}{x+1}=\frac{2-\frac{1-5x}{2}-x}{x+1}=\frac{3x+3}{2(x+1)}=\boxed{\frac{3}{2}}.$$