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8. Sequences and Series
hard
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.
A
$57$
B
$56$
C
$55$
D
$54$
(JEE MAIN-2022)
Solution
${ }^{4} C _{2} \times \frac{\beta^{2}}{6},-6 \beta,-{ }^{6} C _{3} \times \frac{\beta^{3}}{8}$ are in A.P
$\beta^{2}-\frac{5}{2} \beta^{3}=-12 \beta$
$\beta=\frac{12}{5} \text { or } \beta=-2 \therefore \beta=\frac{12}{5}$
$d =-\frac{72}{5}-\frac{144}{25}=-\frac{504}{25}$
$\therefore 50-\frac{2 d }{\beta^{2}}=57$
Standard 11
Mathematics
