JEE Main 20245 Apr, Morning shiftMathematicsMediumSequence and Series179× since 2002If 11+2+12+3+…+199+100=m\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m1+21+2+31+…+99+1001=m and 11⋅2+12⋅3+…+199⋅100=n\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=\mathrm{n}1⋅21+2⋅31+…+99⋅1001=n, then the point (m,n)(\mathrm{m}, \mathrm{n})(m,n) lies on the lineAOption A. 11(x−1)−100y=011(x-1)-100 y=011(x−1)−100y=0BOption B. 11x−100y=011 x-100 y=011x−100y=0COption C. 11(x−1)−100(y−2)=011(x-1)-100(y-2)=011(x−1)−100(y−2)=0DOption D. 11(x−2)−100(y−1)=011(x-2)-100(y-1)=011(x−2)−100(y−1)=0CheckSkip · show solution
