If 1/√1 + √2 + 1/√2 + √3 + .... + 1/√99 + √100 = m and 1/1.2 + 1/2.3 + .... + 1/99.100 = n,

Question

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

(1) \(11(x-1)-100(y-2)=0\)

(2) \(11(x-2)-100(y-1)=0\)

(3) \(11(x-1)-100 y=0\)

(4) \(11 x-100 y=0\)

Answer 1

Correct option is (4) \(11 x-100 y=0\)

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=\mathrm{m}\)

\(\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} \ldots \frac{\sqrt{99}-\sqrt{100}}{-1}=\mathrm{m}\)

\(\sqrt{100}-1=\mathrm{m} \Rightarrow \mathrm{m}=9 \)

\(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots \frac{1}{99 \cdot 100}=\mathrm{n}\)

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

\(1-\frac{1}{100}=\mathrm{n}\)

\( \frac{99}{100}=\mathrm{n} \)

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

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

\(=99-99=0\)