If the 1011$$^{th}$$ term from the end in the binomial expansion of $$\left(\frac{4x}{5} - \frac{5}{2x}\right)^{2022}$$ is 1024 times 1011$$^{th}$$ term from the beginning, then $$32|x|$$ is equal to

Given the binomial expansion of $$\left(\frac{4x}{5} - \frac{5}{2x}\right)^{2022}$$. The 1011th term from the end is 1024 times the 1011th term from the beginning.

Identify the terms.

In an expansion $$(a+b)^n$$ with $$n+1 = 2023$$ terms, the $$k$$th term from the end is the $$(n+2-k)$$th term from the beginning.

1011th term from the end = $$(2022 + 2 - 1011)$$th = 1013th term from the beginning.

Write the general term.

$$T_{r+1} = \binom{2022}{r}\left(\frac{4x}{5}\right)^{2022-r}\left(-\frac{5}{2x}\right)^r$$

1011th from beginning: $$T_{1011}$$ with $$r = 1010$$.

1013th from beginning: $$T_{1013}$$ with $$r = 1012$$.

Compute the ratio $$|T_{1013}|/|T_{1011}|$$.

$$\frac{|T_{1013}|}{|T_{1011}|} = \frac{\binom{2022}{1012}}{\binom{2022}{1010}} \cdot \left(\frac{4|x|}{5}\right)^{2022-1012-(2022-1010)} \cdot \left(\frac{5}{2|x|}\right)^{1012-1010}$$

Since $$1012 + 1010 = 2022$$, we have $$\binom{2022}{1012} = \binom{2022}{1010}$$, so their ratio is 1.

The remaining ratio simplifies to:

$$\left(\frac{4|x|}{5}\right)^{-2} \cdot \left(\frac{5}{2|x|}\right)^{2} = \frac{25}{16x^2} \cdot \frac{25}{4x^2} = \frac{625}{64x^4}$$

Solve for $$|x|$$.

$$\frac{625}{64x^4} = 1024$$

$$x^4 = \frac{625}{64 \times 1024} = \frac{625}{65536}$$

$$|x|^2 = \frac{25}{256}$$

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

Compute $$32|x|$$.

$$32|x| = 32 \times \frac{5}{16} = 10$$

The answer is Option B: $$10$$.