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

JEE · Math · previous-year question

  1. A.$$11(x-1)-100 y=0$$
  2. B.$$11 x-100 y=0$$correct
  3. C.$$11(x-1)-100(y-2)=0$$
  4. D.$$11(x-2)-100(y-1)=0$$

Answer

B. $$11 x-100 y=0$$

Explanation

$$\begin{aligned} & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m \\ & \text { and } \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n \\ & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \\ & =\frac{1}{\sqrt{1}+\sqrt{2}} \times \frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ & \quad+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}} \times \frac{\sqrt{100}-\sqrt{99}}{\sqrt{100}-\sqrt{99}} \end{aligned}$$ $$\begin{aligned} & =\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{100}-\sqrt{99} \\ & =\sqrt{100}-\sqrt{1} \\ & =10-1 \\ & \Rightarrow m=9 \end{aligned}$$ 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 $$11 x-100 y=0$$

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