QJEEM24Ma7533: JEE Main 2024 Mathematics - Class 11 JEEM24Ma7533 Solution: If and , then the point lies on the line

$\begin{aligned}& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\& \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\& \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\& =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\& \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}}\end{aligned}$

$\begin{aligned}& =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\& =\sqrt{100}-\sqrt{1} \\& =10-1 \\& \Rightarrow m=9\end{aligned}$

and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$

$\begin{aligned}& \frac{2-1}{1 \times 2}+\frac{3-2}{2 \times 3}+\ldots+\frac{100-99}{100 \times 99}=n \\& \Rightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\ldots+\frac{1}{99}-\frac{1}{100}=n \\& \Rightarrow n=1-\frac{1}{100} \\& \Rightarrow n=\frac{99}{100} \\& (m, n)=\left(9, \frac{99}{100}\right)\end{aligned}$

Satisfies the line $11 x-100 y=0$