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QJEEM22Ma1096: JEE Main 2022 Mathematics - Class 11 JEEM22Ma1096 Solution: Let the coefficients of the middle terms in the expansion of…

Let the coefficients of the middle terms in the expansion of (16+βx)4,(1−3βx)2\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 50−2dβ250-\frac{2 d}{\beta^{2}} is equal to __________.

Step-by-Step Verified Solution

Coefficients of middle terms of given expansions are 4C216β2, 2C1(−3β), 6C3(−β2)3{}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3} form an A.P.

∴\therefore 2.2(−3β)=β2−5β322.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}

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

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

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

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

β=125\beta = {{12} \over 5}

d=−6β−β2d = - 6\beta - {\beta ^2}

∴\therefore 50−2dβ2=50−2(−6β−β2)β2=50+12β+2=5750 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57

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