Question

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

• A. 11(x – 1) – 100(y – 2) = 0
• B. 11(x – 2) – 100(y – 1) = 0
• C. 11(x – 1) – 100y = 0
• D. 11x – 100y = 0

Answer 1

Answer: (D)

11x – 100y = 0

Answer 2

For $m$:

$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} + \sqrt{100}} = m$

Rationalize each term:

$\frac{\sqrt{1} - \sqrt{2}}{-1} + \frac{\sqrt{2} - \sqrt{3}}{-1} + \cdots + \frac{\sqrt{99} - \sqrt{100}}{-1} = m$

This telescopes to:

$\sqrt{100} - 1 = m \implies m = 9$

For $n$:

$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{99 \cdot 100} = n$

Rewrite as:

$\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{99} - \frac{1}{100}\right) = n$

This telescopes to:

$1 - \frac{1}{100} = n \implies n = \frac{99}{100}$

Thus, $(m, n) = (9, \frac{99}{100})$.

Substitute into the equation:

$11(9) - 100\left(\frac{99}{100}\right) = 99 - 99 = 0$

Correct Ans. option (4) $11x - 100y = 0$