If the middle term in the expansion of `( (2)/(3) x^(2) - (3)/(2x)) ^(20) is^(20) C_10 x^(k)` , then find the value of k .

Answer

To find the value of \( k \) in the expression \( \left( \frac{2}{3} x^2 - \frac{3}{2x} \right)^{20} \), we will follow these steps:

### Step 1: Identify the middle term
The expression is of the form \( (a + b)^n \) where:
- \( a = \frac{2}{3} x^2 \)
- \( b = -\frac{3}{2x} \)
- \( n = 20 \)

Since \( n \) is even, the middle term in the expansion is given by the formula:
\[ \text{Middle term} = T_{\frac{n}{2} + 1} = T_{10 + 1} = T_{11} \]

### Step 2: Find the 11th term
The general term \( T_{r+1} \) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
To find \( T_{11} \), we set \( r = 10 \):
\[ T_{11} = \binom{20}{10} a^{20-10} b^{10} \]

### Step 3: Substitute values of \( a \) and \( b \)
Substituting \( a \) and \( b \):
\[ T_{11} = \binom{20}{10} \left(\frac{2}{3} x^2\right)^{10} \left(-\frac{3}{2x}\right)^{10} \]

### Step 4: Simplify the expression
Calculating \( T_{11} \):
\[ T_{11} = \binom{20}{10} \left(\frac{2^{10}}{3^{10}} x^{20}\right) \left(-\frac{3^{10}}{2^{10} x^{10}}\right) \]
This simplifies to:
\[ T_{11} = \binom{20}{10} \left(\frac{2^{10}}{3^{10}} \cdot -\frac{3^{10}}{2^{10}}\right) x^{20 - 10} \]
\[ = \binom{20}{10} (-1) x^{10} \]

### Step 5: Compare with given expression
We are given that the middle term is \( \binom{20}{10} x^k \). Thus, we can equate:
\[ \binom{20}{10} (-1) x^{10} = \binom{20}{10} x^k \]

### Step 6: Determine the value of \( k \)
From the comparison, we see that:
\[ k = 10 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{10} \]