If the \(1011^{th}\) term from the end in the binominal expansion of \((\frac{4x}{5}-\frac{5}{2x})^{2022}\) is 1024 times \(1011^{th}\) term from the beginning, then |x| is equal to
- 10
- 8
- 15
- 12
The Correct Option is A
Solution and Explanation
Step 1: General term in the binomial expansion
The general term in the expansion of \(\left(4x^5 - \frac{5}{2x}\right)^{2022}\) is given by:
\[ T_r = \binom{2022}{r-1} \cdot (4x^5)^{2022-(r-1)} \cdot \left(-\frac{5}{2x}\right)^{r-1} \]
Step 2: \(1011^{th}\) term from the beginning
The \(1011^{th}\) term from the beginning corresponds to \(r = 1011\):
\[ T_{1011} = \binom{2022}{1010} \cdot (4x^5)^{1012} \cdot \left(-\frac{5}{2x}\right)^{1010} \]
Step 3: \(1011^{th}\) term from the end
The \(1011^{th}\) term from the end corresponds to \(r = 1012\) from the beginning. Substituting \(r = 1012\):
\[ T_{1011}^{\text{end}} = \binom{2022}{1011} \cdot (4x^5)^{1010} \cdot \left(-\frac{5}{2x}\right)^{1012} \]
Step 4: Relationship between the two terms
According to the problem, the term from the end is 1024 times the term from the beginning:
\[ T_{1011}^{\text{end}} = 1024 \cdot T_{1011} \]
Substituting the expressions for the terms:
\[ \binom{2022}{1011} \cdot (4x^5)^{1010} \cdot \left(-\frac{5}{2x}\right)^{1012} = 1024 \cdot \binom{2022}{1010} \cdot (4x^5)^{1012} \cdot \left(-\frac{5}{2x}\right)^{1010} \]
Step 5: Simplify the equation
Cancel common terms and simplify the powers of \(x\):
\[ \frac{\binom{2022}{1011}}{\binom{2022}{1010}} \cdot \frac{(4x^5)^{1010} \cdot \left(-\frac{5}{2x}\right)^{1012}}{(4x^5)^{1012} \cdot \left(-\frac{5}{2x}\right)^{1010}} = 1024 \]
Simplifying further gives:
\[ \frac{1}{4x^2} = 1024 \]
Step 6: Solve for \(|x|\)
Rearrange the equation:
\[ x^2 = \frac{1}{4 \cdot 1024} = \frac{1}{4096} \]
Taking the square root:
\[ |x| = 10 \]
Final Answer
\(|x| = 10\)
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