Let the coefficients of the middle terms in the expansion of ( 1 √6+β x)^4,(1-3β x)^2 and (1- β 2x)^6,β >0, respectively form the first three terms…

Let the coefficients of the middle terms in the expansion of ${(\frac{1}{\sqrt{6}}+\beta x)}^{4},{(1-3\beta x)}^{2}$(6​1​+βx)4,(1−3βx)2 and ${(1-\frac{\beta }{2}x)}^{6},\beta >0$(1−2β​x)6,β>0, respectively form the first three terms of an A.P. If $d$d is the common difference of this A.P., then $50-\frac{2d}{{\beta }^{2}}$50−β22d​ is equal to _____ .

Given,

Coefficients of middle terms of given expansions ${(\frac{1}{\sqrt{6}}+\beta x)}^{4}$(6​1​+βx)4, ${(1-3\beta x)}^{2}$(1−3βx)2 and${(1-\frac{\beta }{2}x)}^{6}$(1−2β​x)6 are C24\frac{1}{6}{\beta }^{2},C12(-3\beta )&C36{(\frac{-\beta }{2})}^{3} respectively,

Now it is given that middle terms are forming an $A.P$A.P

So, $2C12(-3\beta )=C36{(\frac{-\beta }{2})}^{3}+C24\frac{1}{6}{\beta }^{2}$2C12(−3β)=C36(2−β​)3+C2461​β2

$$\Rightarrow 2.2(-3\beta )={\beta }^{2}-\frac{5{\beta }^{3}}{2}$$⇒2.2(−3β)=β2−25β3​

$$\Rightarrow -24=2\beta -5{\beta }^{2}$$⇒−24=2β−5β2

$$\Rightarrow 5{\beta }^{2}-2\beta -24=0$$⇒5β2−2β−24=0

$$\Rightarrow 5{\beta }^{2}-12\beta +10\beta -24=0$$⇒5β2−12β+10β−24=0

$$\Rightarrow \beta (5\beta -12)+2(5\beta -12)=0$$⇒β(5β−12)+2(5β−12)=0

$$\Rightarrow (\beta +2)(5\beta -12)=0$$⇒(β+2)(5β−12)=0

So, $\beta =\frac{12}{5}$β=512​ and $\beta =-2$β=−2 {which is not possible as $\beta >0$β>0}

Now $d=C12(-3\beta )-C24\frac{1}{6}{\beta }^{2}=-6\beta -{\beta }^{2}$d=C12(−3β)−C2461​β2=−6β−β2

Now putting the value of \beta &d in $50-\frac{2d}{{\beta }^{2}}$50−β22d​ we get,

$$50-\frac{2d}{{\beta }^{2}}=50-2\frac{(-6\beta -{\beta }^{2})}{{\beta }^{2}}=50+\frac{12}{\beta }+2=57$$50−β22d​=50−2β2(−6β−β2)​=50+β12​+2=57