Skip to main content

If 1 √1+√2+ 1 √2+√3+…+ 1 √99+√100=m and 1 1 · 2+ 1 2 · 3+…+ 1 99 · 100=n, then the point ( m, n) lies on the line

JEE Main 2024 — Mathematics Algebra

2024mcqmedium

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}=n1⋅21​+2⋅31​+…+99⋅1001​=n, then the point (m,n)(\mathrm{m}, \mathrm{n})(m,n) lies on the line

Official previous-year question

Held on 5 Apr 2024 · Verified 6 Jul 2026.

Options

  1. A

    11(x−1)−100(y−2)=011(x-1)-100(y-2)=011(x−1)−100(y−2)=0

  2. B

    11x−100y=011 x-100 y=011x−100y=0

  3. C

    11(x−2)−100(y−1)=011(x-2)-100(y-1)=011(x−2)−100(y−1)=0

  4. D

    11(x−1)−100y=011(x-1)-100 y=011(x−1)−100y=0

Did you get this right?

Sign in to track your attempts and accuracy.

Your note

Sign in to keep a private note on this question. Nothing you write is ever public.

JEE Main Algebra in other years

More JEE Main Algebra questions

Other Mathematics topics for JEE Main

Keep practicing Algebra

Work through every JEE Main Algebra PYQ, year by year.

Browse all Mathematics topics