JIPMAT 2021 (QA) - The value of (1sqrt(9)-sqrt(8) - 1sqrt(8)-sqrt(7) + 1sqrt(7)-sqrt(6) - 1sqrt(6)-sqrt(5) + 1sqrt(5)-sqrt(4)) is | PYQs + Solutions

JIPMAT 2021

Algebra

Progression & Series

Conceptual

The value of $(\frac{1}{\sqrt{9}-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} - \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-\sqrt{4}})$ is

\frac{1}{5}

Correct Option: 4

Let's look at this sum:

\frac{1}{\sqrt{9}-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} - \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-\sqrt{4}}

Let's rationalize each denominator:

\frac{1}{\sqrt{9}-\sqrt{8}} = \frac{\sqrt{9}+\sqrt{8}}{(\sqrt{9}-\sqrt{8})(\sqrt{9}+\sqrt{8})} = \frac{\sqrt{9}+\sqrt{8}}{9-8} = \frac{3+2\sqrt{2}}{1} = 3+2\sqrt{2}

Similarly,

\newline

\frac{1}{\sqrt{8}-\sqrt{7}} = -(2\sqrt{2}+\sqrt{7})

\frac{1}{\sqrt{7}-\sqrt{6}} = \sqrt{7}+\sqrt{6}

\frac{1}{\sqrt{6}-\sqrt{5}} = -(\sqrt{6}+\sqrt{5})

\frac{1}{\sqrt{5}-\sqrt{4}} = \sqrt{5}+2

Adding all terms:

\newline

3+2\sqrt{2}-(2\sqrt{2}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+2)

All terms cancel except 5.

Therefore, the value is 5.