`overset(2n)underset(r=1)Sigma sin^(-1) x_r=npi " then " overset(2n)underset(r=1)Sigma x_i=`

Options

A. n
B. `2n`
C. `(n(n+1))/2`
D. none

Answer

To solve the problem, we start with the equation given in the question:

\[ \sum_{r=1}^{2n} \sin^{-1}(x_r) = n\pi \]

We need to find the value of

\[ \sum_{i=1}^{2n} x_i \]

### Step 1: Understanding the Equation
The equation states that the sum of the inverse sine (arcsine) of \(x_r\) from \(r=1\) to \(2n\) equals \(n\pi\). The arcsine function, \(\sin^{-1}(x)\), has a range of \([- \frac{\pi}{2}, \frac{\pi}{2}]\).

### Step 2: Analyzing the Values of \(x_r\)
Since the sum of arcsines equals \(n\pi\), we can infer that each term \(\sin^{-1}(x_r)\) must be equal to \(\frac{\pi}{2}\) for \(n\) terms, and the remaining \(n\) terms must be equal to \(-\frac{\pi}{2}\) to balance the equation.

This leads us to conclude that:
- For \(n\) values of \(r\), \(x_r = 1\) (since \(\sin^{-1}(1) = \frac{\pi}{2}\)).
- For \(n\) values of \(r\), \(x_r = -1\) (since \(\sin^{-1}(-1) = -\frac{\pi}{2}\)).

### Step 3: Summing the Values of \(x_r\)
Now we can sum the values of \(x_r\):

\[ \sum_{r=1}^{2n} x_r = \sum_{r=1}^{n} 1 + \sum_{r=1}^{n} (-1) \]

This simplifies to:

\[ \sum_{r=1}^{2n} x_r = n \cdot 1 + n \cdot (-1) = n - n = 0 \]

### Conclusion
Thus, the value of

\[ \sum_{i=1}^{2n} x_i = 0 \]