JEE Mains · Maths · STD 11 - 8. sequence and series
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
- A \(11(x-1)-100(y-2)=0\)
- B \(11(x-2)-100(y-1)=0\)
- C \(11(x-1)-100 y=0\)
- D \(11 x-100 y=0\)
Answer & Solution
Correct Answer
(D) \(11 x-100 y=0\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=\mathrm{m} \) \( \frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} \ldots \frac{\sqrt{99}-\sqrt{100}}{-1}=\mathrm{m}\) \( \sqrt{100}-1=\mathrm{m} \Rightarrow \mathrm{m}=9 \)…
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