JEE Main 2024 — 5 April, Shift 1 — Question 16

$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$

$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} \ldots \frac{\sqrt{99}-\sqrt{100}}{-1}=m$

$\sqrt{100}-1=m \Rightarrow \mathrm{~m}=9$

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

$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} \ldots \frac{1}{99}-\frac{1}{100}=\mathrm{n}$

$1-\frac{1}{100}=\mathrm{n}$

$\frac{99}{100}=\mathrm{n}$

$(m, n)=\left(9, \frac{99}{100}\right)$

$\Rightarrow 11(9)-100\left(\frac{99}{100}\right)$

$=99-99=0$

$11 x-100 y=0$