If the $1011^{\text {th }}$ term from the end in the binominal expansion of $\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022}$ is 1024 times $1011^{\text {th }}$R term from the beginning, then $|x|$ is equal to

$\mathrm{T}_{1011}$ from beginning $=\mathrm{T}_{1010+1}$ <br/><br/>$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010} $$ <br/><br/>$\mathrm{T}_{1011}$ from end <br/><br/>$$ ={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} $$ <br/><br/>$$ \begin{aligned} & \text { Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\ & =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012} \end{aligned} $$ <br/><br/>$$ \begin{aligned} &\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\ &\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\ & \Rightarrow |x|=\frac{5}{16} \end{aligned} $$