Let the coefficients of the middle terms in the expansion of ${\left(\frac{1}{\sqrt{6}}+\beta x\right)}^4, {\left(1-3\beta x\right)}^2$

and ${\left(1-\frac{\beta }{2}x\right)}^6, \beta >0$, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then $50-\frac{2d}{{\beta }^2}$ is equal to ______

$$\begin{gathered} Coefficient of middle term in {\left(\frac{1}{\sqrt{6}}+\beta x\right)}^4 is{ }^4{C}_2\times \frac{{\beta }^2}{6}, \\ Coefficient of middle term in {\left(1-3\beta x\right)}^2 is -6\beta \\ and Coefficient of middle term in{\left(1-\frac{\beta }{2}x\right)}^6 is -{}^6{C}_3\times \frac{{\beta }^3}{8} \\ These coefficients are in A.P. \\ {\Rightarrow }^4{C}_2\times \frac{{\beta }^2}{6}, -6\beta , -{}^6{C}_3\times \frac{{\beta }^3}{8} are in A.P \\ \therefore 2\left(-6\beta \right){=}^4{C}_2\times \frac{{\beta }^2}{6}+\left(-{}^6{C}_3\times \frac{{\beta }^3}{8}\right) \\ \Rightarrow 6\frac{{\beta }^2}{6}-20\frac{{\beta }^3}{8}=-12\beta \\ \Rightarrow {\beta }^2-\frac{5}{2}{\beta }^3=-12\beta \\ \Rightarrow \frac{5}{2}{\beta }^2-\beta -12=0 \\ \beta =\frac{12}{5} or \beta =-2 \therefore \beta =\frac{12}{5}\sin ce\beta >0 \\ {\beta }^2,-6\beta ,\frac{-5}{2}{\beta }^{3}are in A.P \\ \Rightarrow \frac{144}{25},-\frac{72}{5},\frac{-864}{25}are in A.P \\ d=-\frac{72}{5}-\frac{144}{25}=-\frac{504}{25} \\ \therefore 50-\frac{2d}{{\beta }^2}=50-2\left(\frac{-504}{25}\right)\frac{25}{144}=50+\frac{1008}{144}=57 \end{gathered}$$