If ` Sigma_( i = 1)^( 2n) sin^(-1) x_(i) = n pi` , then find the value of `Sigma_( i = 1)^( 2n) x_(i)`.

Answer

To solve the problem, we need to find the value of \( \Sigma_{i=1}^{2n} x_i \) given that \( \Sigma_{i=1}^{2n} \sin^{-1}(x_i) = n\pi \).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
We start with the equation:
\[ \Sigma_{i=1}^{2n} \sin^{-1}(x_i) = n\pi \]
The range of the function \( \sin^{-1}(x) \) is from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). Therefore, for \( 2n \) terms, the maximum possible sum of \( \sin^{-1}(x_i) \) occurs when each \( \sin^{-1}(x_i) = \frac{\pi}{2} \).

2. **Finding the Maximum Value**:
If each \( \sin^{-1}(x_i) \) were to equal \( \frac{\pi}{2} \), then:
\[ \Sigma_{i=1}^{2n} \sin^{-1}(x_i) = 2n \cdot \frac{\pi}{2} = n\pi \]
This indicates that each \( \sin^{-1}(x_i) \) must equal \( \frac{\pi}{2} \) for the sum to equal \( n\pi \).

3. **Determining Values of \( x_i \)**:
Since \( \sin^{-1}(x_i) = \frac{\pi}{2} \), we have:
\[ x_i = \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus, all \( x_i \) values must be equal to 1.

4. **Calculating the Sum**:
Now, we can compute the sum:
\[ \Sigma_{i=1}^{2n} x_i = x_1 + x_2 + \ldots + x_{2n} = 1 + 1 + \ldots + 1 \quad (2n \text{ times}) \]
This simplifies to:
\[ \Sigma_{i=1}^{2n} x_i = 2n \]

5. **Final Answer**:
Therefore, the value of \( \Sigma_{i=1}^{2n} x_i \) is:
\[ \boxed{2n} \]