Question:
The constant term in $\left(\frac{\sqrt{x}}{2} + \frac{1}{3x^2}\right)^{10}$ is:
The constant term in $\left(\frac{\sqrt{x}}{2} + \frac{1}{3x^2}\right)^{10}$ is:
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To find the constant term, set the total exponent of $x$ to zero.
Updated On: Apr 28, 2026
- $\frac{5}{128}$
- $\frac{9}{128}$
- $\frac{5}{256}$
- $\frac{9}{256}$
- 0
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The Correct Option is C
Solution and Explanation
Step 1: Formula
General term $T_{r+1} = {}^{10}C_r (\frac{x^{1/2}}{2})^{10-r} (\frac{1}{3x^2})^r$.
Step 2: Analysis
Power of $x = \frac{10-r}{2} - 2r$. For constant term, power $= 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$.
Step 3: Calculation
$T_3 = {}^{10}C_2 (\frac{1}{2})^8 (\frac{1}{3})^2 = 45 \cdot \frac{1}{256} \cdot \frac{1}{9} = \frac{5}{256}$. Final Answer: (C)
General term $T_{r+1} = {}^{10}C_r (\frac{x^{1/2}}{2})^{10-r} (\frac{1}{3x^2})^r$.
Step 2: Analysis
Power of $x = \frac{10-r}{2} - 2r$. For constant term, power $= 0 \implies 10 - r - 4r = 0 \implies 5r = 10 \implies r = 2$.
Step 3: Calculation
$T_3 = {}^{10}C_2 (\frac{1}{2})^8 (\frac{1}{3})^2 = 45 \cdot \frac{1}{256} \cdot \frac{1}{9} = \frac{5}{256}$. Final Answer: (C)
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