In a triangle $ABC$, the measure of angles $A,B,C$ are in $AP$ and $b:c=\sqrt{3}:\sqrt{2}$ then the measure of angle $A$ is $\frac{5\mathrm{\pi }}{2k}$ then the value of $k$ is 

Given that $A,B.C$ are in $AP$

$2B=A+C$

Hence, $B=60°$

GIven $b:c=\sqrt{3}:\sqrt{2}$

Use sine rule 

$2R \sin B:2R\sin C=\sqrt{3}:\sqrt{2}$

$\begin{array}{rcl}\frac{\sin B}{\sin C}& =& \frac{\sqrt{3}}{\sqrt{2}}\\ \frac{\sin 60}{\sin C}& =& \frac{\sqrt{3}}{\sqrt{2}}\\ \sin C& =& \frac{1}{\sqrt{2}}\\ C& =& 45°\end{array}$

Hence, $A=180°-45°-60°=105°$

Therefore, $A=\frac{7\mathrm{\pi }}{12}=\frac{7\mathrm{\pi }}{2k}$ it implies that $k=\boxed{\mathbf{6}}$