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 [2024]

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Solution

Q.
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 [2024]

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

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

3 $11 x - 100 y = 0$

4 $11 (x - 1) - 100 y = 0$

Ans.
(3)

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

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

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

$\Rightarrow \sqrt{100} - 1 = m \Rightarrow m = 10 - 1 = 9 \dots (i)$

Also, $(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{99} - \frac{1}{100}) = n$

$\Rightarrow 1 - \frac{1}{100} = n \Rightarrow n = \frac{99}{100} \dots (i i)$

$∴$ $(m, n) = (9, \frac{99}{100})$

So, $(m, n)$ lies on $11 x - 100 y = 0$

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