1+21+2+31+…+99+1001=m−11−2+−12−3…−199−100=m $\begin{aligned}
& \sqrt{100}-1=\mathrm{m} \Rightarrow \mathrm{m}=9 \
& \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots \frac{1}{99 \cdot 100}=\mathrm{n} \
& \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} \ldots \frac{1}{99}-\frac{1}{100}=\mathrm{n} \
& 1-\frac{1}{100}=\mathrm{n} \
& \frac{99}{100}=\mathrm{n} \
& (\mathrm{m}, \mathrm{n})=\left(9, \frac{99}{100}\right) \
& \Rightarrow 11(9)-100\left(\frac{99}{100}\right) \
& =99-99=0
\end{aligned}Ans.11 x-100 y=0$
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
If 1+21+2+31+…+99+1001=m and 1⋅21+2⋅31+…+99⋅1001=n, then the point (m,n) lies on the line
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
11(x−1)−100(y−2)=0
- B
11x−100y=0
- C
11(x−2)−100(y−1)=0
- D
11(x−1)−100y=0
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