If the 1011^ th term from the end in the binomial expansion of ( 4x 5- 5 2x)^2022 is 1024 times 1011^ th term from the beginning, then 32|x| is equal…

Given,

The ${1011}^{\mathrm{th}}$1011th term from the end in the binomial expansion of ${(\frac{4x}{5}-\frac{5}{2x})}^{2022}$(54x​−2x5​)2022 is $1024$1024 times ${1011}^{\mathrm{th}}$1011th term from the beginning,

So, according to question we have, $\frac{{T}_{1013}}{{T}_{1011}}=1024$T1011​T1013​​=1024

$$\Rightarrow \frac{C10122022{(\frac{4x}{5})}^{1010}\cdot {(\frac{-5}{2x})}^{1012}}{C10102022{(\frac{4x}{5})}^{1012}{(\frac{-5}{2x})}^{1010}}=1024$$⇒C10102022(54x​)1012(2x−5​)1010C10122022(54x​)1010⋅(2x−5​)1012​=1024

$$\Rightarrow \frac{{(\frac{-5}{2x})}^{2}}{{(\frac{4x}{5})}^{2}}=1024$$⇒(54x​)2(2x−5​)2​=1024

$$\Rightarrow \frac{{5}^{4}}{{2}^{6}{x}^{4}}={2}^{10}$$⇒26x454​=210

$$\Rightarrow {(\frac{5}{x})}^{4}={2}^{16}$$⇒(x5​)4=216

$$\Rightarrow |x|=\frac{5}{16}$$⇒∣x∣=165​

$$\Rightarrow 32|x|=10$$⇒32∣x∣=10

Note this question was bonus in Jee Main 2023 April session as answer key was not matching, so we have modified the question.