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Sequences and Series question

2024 · 5 Apr · Shift 1 · Q42
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  5. /2024 · 5 Apr · Shift 1 · Q42

Sequences and Series question

2024 · 5 Apr · Shift 1 · Q42

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If 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​+2​1​+2​+3​1​+…+99​+100​1​=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 line
  1. A
    11(x−1)−100y=011(x-1)-100 y=011(x−1)−100y=0
  2. B
    11x−100y=011 x-100 y=011x−100y=0
  3. C
    11(x−1)−100(y−2)=011(x-1)-100(y-2)=011(x−1)−100(y−2)=0
  4. D
    11(x−2)−100(y−1)=011(x-2)-100(y-1)=011(x−2)−100(y−1)=0
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