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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
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
    $11(x-1)-100 y=0$
  • ✓
    $11 x-100 y=0$
✓

Answer

Correct option: D.
$11 x-100 y=0$
d
$ \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 $

$ \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$

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

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