Let the middle-term coefficients of the three given expansions be the first three terms of an A.P.

We compute them one by one.

1. Coefficient of the middle term in

$\left(\frac{1}{\sqrt{6}}+\beta x\right)^4$

For $(a+b)^4$, the single middle term is the $\left(\frac{4}{2}+1\right)=3$rd term.

General term: $T_{r+1}=\binom{4}{r}\left(\frac{1}{\sqrt{6}}\right)^{4-r}(\beta x)^r$

For the middle term, $r=2$: $T_3=\binom{4}{2}\left(\frac{1}{\sqrt{6}}\right)^2(\beta x)^2$

So its coefficient is $\binom{4}{2}\cdot \frac{1}{6}\cdot \beta^2=6\cdot \frac{1}{6}\beta^2=\beta^2$

Thus, $a_1=\beta^2$

2. Coefficient of the middle term in

$(1-3\beta x)^2$

For power $2$, the single middle term is the 2nd term.

Expansion: $(1-3\beta x)^2=1-6\beta x+9\beta^2x^2$

So the middle term is $-6\beta x$, whose coefficient is $a_2=-6\beta$

3. Coefficient of the middle term in

$\left(1-\frac{\beta}{2}x\right)^6$

For $(a+b)^6$, the single middle term is the $\left(\frac{6}{2}+1\right)=4$th term.

General term: $T_{r+1}=\binom{6}{r}(1)^{6-r}\left(-\frac{\beta}{2}x\right)^r$

For the middle term, $r=3$: $T_4=\binom{6}{3}\left(-\frac{\beta}{2}x\right)^3$

Therefore coefficient is $a_3=\binom{6}{3}\left(-\frac{\beta}{2}\right)^3=20\left(-\frac{\beta^3}{8}\right)=-\frac{5}{2}\beta^3$

4. Use A.P. condition

Since $a_1,a_2,a_3$ are in A.P., $2a_2=a_1+a_3$

Substitute: $2(-6\beta)=\beta^2-\frac{5}{2}\beta^3$

$-12\beta=\beta^2-\frac{5}{2}\beta^3$

Since $\beta>0$, divide by $\beta$: $-12=\beta-\frac{5}{2}\beta^2$

Multiply by $2$: $-24=2\beta-5\beta^2$

$5\beta^2-2\beta-24=0$

Solve: $\beta=\frac{2\pm\sqrt{4+480}}{10}=\frac{2\pm 22}{10}$

So, $\beta=\frac{24}{10}=\frac{12}{5} \quad \text{or} \quad \beta=-2$

Given $\beta>0$, hence $\beta=\frac{12}{5}$

5. Find common difference $d$

$d=a_2-a_1=-6\beta-\beta^2$

Substitute $\beta=\frac{12}{5}$: $d=-6\cdot \frac{12}{5}-\left(\frac{12}{5}\right)^2$

$d=-\frac{72}{5}-\frac{144}{25}=-\frac{360+144}{25}=-\frac{504}{25}$

6. Compute the required value

We need $50-\frac{2d}{\beta^2}$

Now, $\beta^2=\frac{144}{25}$

and $2d=2\left(-\frac{504}{25}\right)=-\frac{1008}{25}$

Thus, $\frac{2d}{\beta^2}=\frac{-1008/25}{144/25}=\frac{-1008}{144}=-7$

Therefore, $50-\frac{2d}{\beta^2}=50-(-7)=57$

Final Answer

$\boxed{57}$