If x, (3)/(2), z are in AP; x, 3, z are in GP; then which one of...

$\mathrm{x}, \frac{3}{2}, \mathrm{z} \text{\text{ are }\text{ in } A.P }$$If$a, b, care in $A.P 2 b=a+c$$\begin{aligned} & \therefore 2\left(\frac{3}{2}\right)=\mathrm{x}+\mathrm{z} \\ & \Rightarrow 3=\mathrm{x}+\mathrm{z}...(1) \\ & \mathrm{x}, 3, \mathrm{z} \text{ are in G.P. } \end{aligned}$If $a, b, c$ are in G.P. $b^2=\mathrm{ac}\begin{aligned} & \therefore 3^2=\mathrm{xz} \\ & \Rightarrow 9=\mathrm{xz}...(2) \end{aligned}$ If $x, 6, z$ are in H.P. $\frac{2}{6}=\frac{1}{x}+\frac{1}{z}\begin{aligned} & \qquad\left(\because \frac{2}{b}=\frac{1}{a}+\frac{1}{c} \text{, if } a, b, c \text{ are is H.P. }\right) \\ & \Rightarrow \frac{1}{3}=\frac{z+x}{x z}=\frac{3}{9}=\frac{1}{3}(\operatorname{from}(1)(2)) \\ & \text{ L.H.S. }=\text{ R.H.S. } \end{aligned}$$$