If $\frac{1}{\sqrt{b}+\sqrt{c}},\frac{1}{\sqrt{c}+\sqrt{a}},\frac{1}{\sqrt{a}+\sqrt{b}}$ are in A.P.  then $9^{ax+1},9^{bx+1},9^{cx+1},x\ne 0$ are in:

$\begin{array}{r}\frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}}\\ =\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})}\end{array}$

$\begin{array}{l}\Rightarrow 2\sqrt{ab}+2b+2\sqrt{ac}+2\sqrt{bc}\\ =2\sqrt{bc}+2\sqrt{ac}+c+2\sqrt{ab}+a\\ \Rightarrow 2b=a+c\end{array}$

$$\begin{gathered} \therefore \mathrm{a},\mathrm{b},\mathrm{c},\text{ are in A.P. } \\ \Rightarrow \mathrm{ax},\mathrm{bx},\mathrm{cx}\text{, are in A.P. } \\ \Rightarrow ax+1,bx+1,cx+1\text{, are in A.P. } \\ \Rightarrow ax+1,bx+1,cx+1,\text{ are in A.P. } \\ \Rightarrow 9^{\mathrm{ax}+1},9^{\mathrm{bx}+1},9^{\mathrm{cx}+1}\text{ are in G.P. } \end{gathered}$$

[See the properites of A.P & GP.]