If $\sum\limits^{2n}_{r=1}sin^{-1}x_r = n\, \, \pi, $ then $\sum\limits^{2n}_{r=1} x_r $ is equal to

2n

We have , 

$-\frac{\pi}{2} ≤ sin^{-1} x_i  ≤ \frac{\pi}{2}$ for i = 1, 2...., 2n

$∴ \sum\limits^{2n}_{r=1}sin^{-1}x_r = n\, \, \pi, $

$ ⇒ sin^{-1} x_r =\frac{\pi}{2}; r = 1, 2, ...., 2n$

$⇒ x_r = 1; r = 1, 2, ..., 2n$

$⇒ \sum\limits^{2n}_{r=1} x_r = \sum\limits^{2n}_{r=1} 1 = 2n $