Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ 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 $50-\frac{2 d}{\beta^{2}}$ is equal to __________.

<p>Coefficients of middle terms of given expansions are $${}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3}$$ form an A.P.</p> <p>$\therefore$ $2.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}$</p> <p>$\Rightarrow - 24 = 2\beta - 5{\beta ^2}$</p> <p>$\Rightarrow 5{\beta ^2} - 2\beta - 24 = 0$</p> <p>$\Rightarrow 5{\beta ^2} - 12\beta + 10\beta - 24 = 0$</p> <p>$\Rightarrow \beta (5\beta - 12) + 2(5\beta - 12) = 0$</p> <p>$\beta = {{12} \over 5}$</p> <p>$d = - 6\beta - {\beta ^2}$</p> <p>$\therefore$ $$50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$$</p>