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

Correct answer: B. 11 x-100 y=0 aligned & 1 1 + 2 + 1 2 + 3 + + 1 99 + 100 =m & 1 - 2 -1 + 2 - 3 -1 99 - 100 -1 =m aligned aligned & 100 -1=m m=9 & (1)/(1 · 2)+(1)/(2 · 3)+ (1)/(99 · 100)=n & (1)/(1)-(1)/(2)+(1)/(2)-(1)/(3) (1)/(99)-(1)/(100)=n & 1-(1)/(100)=n & (99)/(100)=n & (m, n)=(9, (99)/(100)) & 11(9)-100((99)/(100)) & =99-99=0 aligned Ans. 11 x-100 y=0

JEE Main 20245 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Sequences and Series

If 1 1 + 2 + 1 2 + 3 + + 1 99 + 100 =m and (1)/(1 · 2)+(1)/(2 · 3)+...

\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}=n

, then the point

(\mathrm{m}, \mathrm{n})

lies on the line

Options

A $11(x-1)-100(y-2)=0$
B $11 x-100 y=0$
C $11(x-2)-100(y-1)=0$
D $11(x-1)-100 y=0$

Correct answer

B. $11 x-100 y=0$

Step-by-step solution

\begin{aligned} & \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}\end{aligned}

\begin{aligned} & \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 \end{aligned}

Ans.

11 x-100 y=0$

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