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

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Binomial Theorem

If the

101 1^{th}

term from the beginning in the binomial expansion of

(\frac{4 x}{5} - \frac{5}{2 x})^{2022}

is 1024 times the

101 1^{th}

term from the end, then

∣ x ∣

is equal to

Answer: (C)

There are 2023 terms. The

101 1^{th}

term from the beginning is

T_{1011} = (1010 2022) (\frac{4 x}{5})^{1012} (- \frac{5}{2 x})^{1010}

; the

101 1^{th}

from the end is the

101 3^{th}

from the beginning,

T_{1013} = (1012 2022) (\frac{4 x}{5})^{1010} (- \frac{5}{2 x})^{1012}

. As

(1012 2022) = (1010 2022)

\frac{T _{1011}}{T _{1013}} = (\frac{4 x}{5})^{2} (- \frac{5}{2 x})^{- 2} = \frac{64 x ^{4}}{625}

. Setting this

= 1024

gives

x^{4} = \frac{1024 \cdot 625}{64} = 10000

, so

∣ x ∣ = 10

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option