Question

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 __

Answer 1

${ }^4 C_2 \times \frac{\beta^2}{6},-6 \beta,-{ }^6 C_3 \times \frac{\beta^3}{8}$ are in A.P
$ \beta^2-\frac{5}{2} \beta^3=-12 \beta$
$ \beta=\frac{12}{5} \text { or } \beta=-2$
$ \therefore \beta=\frac{12}{5}$
$ d=-\frac{72}{5}-\frac{144}{25}=-\frac{504}{25}$
$ \therefore 50-\frac{2 d}{\beta^2}=57$