Let the coefficients of the middle terms in the expansion of ( 1 6...

Let the coefficients of the middle terms in the expansion of ${\left(\dfrac{1}{\sqrt{6}} + \betax\right)}^{4}, {\left(1 - 3\betax\right)}^{2} and${\left(1 - \dfrac{\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$50 - \dfrac{2d}{{\beta}^{2}}is equal to _____ .</p><p>Given,</p><p>Coefficients of middle terms of given expansions{\left(\dfrac{1}{\sqrt{6}} + \betax\right)}^{4}$,${\left(1 - 3\betax\right)}^{2} and ${\left(1 - \dfrac{\beta}{2}x\right)}^{6} are$C2 4 \dfrac{1}{6}{\beta}^{2}, C1 2 \left(- 3\beta\right) \text{ and } C3 6 {\left(\dfrac{- \beta}{2}\right)}^{3}$respectively,</p><p>Now it is given that middle terms are forming an$A . P$</p><p>So,$2 C1 2 \left(- 3\beta\right) = C3 6 {\left(\dfrac{- \beta}{2}\right)}^{3} + C2 4 \dfrac{1}{6}{\beta}^{2}$</p><p>$⇒ 2 . 2\left(- 3\beta\right) = {\beta}^{2} - \dfrac{5{\beta}^{3}}{2}$</p><p>$⇒ - 24 = 2\beta - 5{\beta}^{2}$</p><p>$⇒ 5{\beta}^{2} - 2\beta - 24 = 0$</p><p>$⇒ 5{\beta}^{2} - 12\beta + 10\beta - 24 = 0$</p><p>$⇒ \beta\left(5\beta - 12\right) + 2\left(5\beta - 12\right) = 0$</p><p>$⇒ \left(\beta + 2\right)\left(5\beta - 12\right) = 0$</p><p>So,$\beta = \dfrac{12}{5}$and$\beta = - 2${which is not possible as \beta > 0}$

Now $d = C1 2 \left(- 3\beta\right) - C2 4 \dfrac{1}{6}{\beta}^{2} = - 6\beta - {\beta}^{2}$

Now putting the value of $\beta \text{ and } d$ in $50 - \dfrac{2d}{{\beta}^{2}}$ we get,

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