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Mathematics · 2024

JEE Main · 5 April 2024, Shift 1 · Q6

If 1/(√ 1 +√ 2)+1/(√ 2 +√ 3)+…+1/(√ 99 +√ 100)=m and 1/(1 · 2)+1/(2 · 3)+…+1/(99 · 100)=n, then the point (m, n) lies on the line

If $\displaystyle \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}+\sqrt{100}}=m$ and $\displaystyle \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots+\frac{1}{99 \cdot 100}=n$, then the point (m, n) lies on the line
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.