JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 15)
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
57
Explanation
Coefficients of middle terms of given expansions are $${}^4{C_2}{1 \over 6}{\beta ^2},\,{}^2{C_1}( - 3\beta ),\,{}^6{C_3}{\left( {{{ - \beta } \over 2}} \right)^3}$$ form an A.P.
$$\therefore$$ $$2.2( - 3\beta ) = {\beta ^2} - {{5{\beta ^3}} \over 2}$$
$$ \Rightarrow - 24 = 2\beta - 5{\beta ^2}$$
$$ \Rightarrow 5{\beta ^2} - 2\beta - 24 = 0$$
$$ \Rightarrow 5{\beta ^2} - 12\beta + 10\beta - 24 = 0$$
$$ \Rightarrow \beta (5\beta - 12) + 2(5\beta - 12) = 0$$
$$\beta = {{12} \over 5}$$
$$d = - 6\beta - {\beta ^2}$$
$$\therefore$$ $$50 - {{2d} \over {{\beta ^2}}} = 50 - 2{{( - 6\beta - {\beta ^2})} \over {{\beta ^2}}} = 50 + {{12} \over \beta } + 2 = 57$$
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