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 _____ .

Given,

Coefficients of middle terms of given expansions ${\left(\frac{1}{\sqrt{6}}+\beta x\right)}^4$, ${\left(1-3\beta x\right)}^2$ and${\left(1-\frac{\beta }{2}x\right)}^6$ are ${}^{4}C_2\frac{1}{6}{\beta }^2, {}^{2}C_1\left(-3\beta \right) \& {}^{6}C_3{\left(\frac{-\beta }{2}\right)}^3$ respectively,

Now it is given that middle terms are forming an $A.P$

So, $2{}^{2}C_1\left(-3\beta \right) = {}^{6}C_3{\left(\frac{-\beta }{2}\right)}^3+{}^{4}C_2\frac{1}{6}{\beta }^2$

$\Rightarrow 2.2\left(-3\beta \right)={\beta }^2-\frac{5{\beta }^3}{2}$

$\Rightarrow -24=2\beta -5{\beta }^2$

$\Rightarrow 5{\beta }^2-2\beta -24=0$

$\Rightarrow 5{\beta }^2-12\beta +10\beta -24=0$

$\Rightarrow \beta \left(5\beta -12\right)+2\left(5\beta -12\right)=0$

$\Rightarrow \left(\beta +2\right)\left(5\beta -12\right)=0$

So, $\beta =\frac{12}{5}$ and $\beta =-2$ {which is not possible as $\beta >0$}

Now $d={}^{2}C_1\left(-3\beta \right)-{}^{4}C_2\frac{1}{6}{\beta }^2=-6\beta -{\beta }^2$

Now putting the value of $\beta \& d$ in $50-\frac{2d}{{\beta }^2}$ we get,

$50-\frac{2d}{{\beta }^2}=50-2\frac{\left(-6\beta -{\beta }^2\right)}{{\beta }^2}=50+\frac{12}{\beta }+2=57$