(a) Since, $- \frac{\pi}{2} < \sin^{- 1}{x \leq \frac{\pi}{2}}$ $\therefore\sin^{- 1}{x_{i} = \frac{\pi}{2},1 \leq i \leq 20}$ $\Rightarrow x_{i} = 1,\ 1 \leq i \leq 20$ Thus, $\sum_{i = 1}^{20}{x_{i} = 20}$
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