If \[\sum\limits_{i=1}^{2n}{{{\cos }^{-1}}{{x}_{i}}=0}\] then \[\sum\limits_{i=1}^{2n}{{{x}_{i}}}\]is

\[\frac{n\left( n+1 \right)}{2}\]

If \[\sum\limits_{i=1}^{2n}{{{\cos }^{-1}}{{x}_{i}}=0}\] then \[\sum\l

1.	If \[\sum\limits_{i=1}^{2n}{{{\cos }^{-1}}{{x}_{i}}=0}\] then \[\sum\limits_{i=1}^{2n}{{{x}_{i}}}\]is
A.	n
B.	2n
C.	\[\frac{n\left( n+1 \right)}{2}\]
D.	None of these
Answer» C. \[\frac{n\left( n+1 \right)}{2}\]

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If \[\sum\limits_{i=1}^{2n}{{{\cos }^{-1}}{{x}_{i}}=0}\] then \[\sum\limits_{i=1}^{2n}{{{x}_{i}}}\]is

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