Question

If$n\in N,\sum _{k=1}^{n}Si{n}^{-1}\left(x_k\right)=\frac{n\pi }{2}then\sum _{k=1}^n{x}_k=$

• A. $n$
• B. $k$
• C. $\frac{k\left(k+1\right)}{2}$
• D. $\frac{n\left(n+1\right)}{2}$

Answer 1

The correct option is A $n$

${\sum }_{k=1}^{n}si{n}^{-1}\left(x_k\right)=\left(\frac{n\pi }{2}\right)\Rightarrow {\sin }^{-1}{x}_1+si{n}^{-1}{x}_2+.........+si{n}^{-1}{x}_n=\left(\frac{n\pi }{2}\right)asmeanvalueofsi{n}^{-1}x=\left(\frac{\pi }{2}\right)\therefore forLHS=RHSvalueofeachsininversetermshouldbeequalto\left(\frac{\pi }{2}\right)nowsi{n}^{-1}{x}_1=si{n}^{-1}{x}_2=.........=si{n}^{-1}{x}_n=\left(\frac{\pi }{2}\right)\therefore x_1=1,x_2=1,x_3=1,.........,x_n=1then{\sum }_{k=1}^n{x}_k={\sum }_{k=1}^{n}1=n$