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JEE Mains · Maths · STD 11 - 7. binomial theoram

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

  1. A \(12\)
  2. B \(8\)
  3. C \(\frac{5}{16}\)
  4. D \(15\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{5}{16}\)

Step-by-step Solution

Detailed explanation

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 :…
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