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If $n \in N, \sum_{k = 1}^{n} S i n^{- 1} (x_{k}) = \frac{n π}{2}$ then $\sum_{k = 1}^{n} (x_{k}) =$

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Correct option is A

put,∑sin-1k=1nxk=nπ2⇒sin-1xk=π2 for k=1,2,....n ⇒xk=sinπ2 for k=1,2,....n ⇒x1=x2=x3=...........=1∴∑k=1nxk=1+1+1+............+1(n times)=n

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If n∈N,∑k=1nSin−1(xk)=nπ2 then∑k=1n(xk)=

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If $n \in N, \sum_{k = 1}^{n} S i n^{- 1} (x_{k}) = \frac{n π}{2}$ then $\sum_{k = 1}^{n} (x_{k}) =$