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 : ${ }^{2022} C _{\text {1010 }}\left(\frac{-5}{2 x }\right)^{1012}\left(\frac{4 x }{5}\right)^{1010}$

$=2^{10} \cdot{ }^{2022} C _{1000}\left(\frac{-5}{2 x }\right)^{1010}\left(\frac{4 x }{5}\right)^{1012}$

$\left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2$

$x^4=\frac{5^4}{2^{16}}$

$|x|=\frac{5}{16}$