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

If $\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$, then the point $(\mathrm{m}, \mathrm{n})$ lies on the line
  1. $11(x-1)-100(y-2)=0$
  2. $11 x-100 y=0$
  3. $11(x-2)-100(y-1)=0$
  4. $11(x-1)-100 y=0$

Solution

$\begin{aligned} & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=\mathrm{m} \\ & \frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} \ldots \frac{\sqrt{99}-\sqrt{100}}{-1}=\mathrm{m}\end{aligned}$ $\begin{aligned} & \sqrt{100}-1=\mathrm{m} \Rightarrow \mathrm{m}=9 \\ & \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots \frac{1}{99 \cdot 100}=\mathrm{n} \\ & \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} \ldots \frac{1}{99}-\frac{1}{100}=\mathrm{n} \\ & 1-\frac{1}{100}=\mathrm{n} \\ & \frac{99}{100}=\mathrm{n} \\ & (\mathrm{m}, \mathrm{n})=\left(9, \frac{99}{100}\right) \\ & \Rightarrow 11(9)-100\left(\frac{99}{100}\right) \\ & =99-99=0 \end{aligned}$ Ans.$11 x-100 y=0$

Asked in: JEE Main 2024 (05 Apr Shift 1)

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