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If the $1011^{\text {th }}$ term from the end in the binomial expansion of $\left(\frac{4 x}{5}-\frac{5}{2 x }\right)^{2022}$ is $1024$ times $1011^{\text {th }}$ term from the beginning, then $|x|$ is equal to
$12$
$8$
$\frac{5}{16}$
$15$
Solution
Sol. $T _{1011}$ from beginning $= T _{1010+1}$
$={ }^{2022} C_{1010}\left(\frac{4 x}{5}\right)^{1012}\left(\frac{-5}{2 x }\right)^{1010}$
$T _{1011}$ from end
$={ }^{2022} C_{1010}\left(\frac{-5}{2 x }\right)^{1012}\left(\frac{4 x }{5}\right)^{1010}$
Given : ${ }^{2022} C _{\text {1010 }}\left(\frac{-5}{2 x }\right)^{1012}\left(\frac{4 x }{5}\right)^{1010}$
$=2^{10} \cdot{ }^{2022} C _{1000}\left(\frac{-5}{2 x }\right)^{1010}\left(\frac{4 x }{5}\right)^{1012}$
$\left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2$
$x^4=\frac{5^4}{2^{16}}$
$|x|=\frac{5}{16}$
