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SSC CHSL 4 Mar 2018 Shift 3 Paper

Section: English Language

Question : 14 of 100

Marks: +1, -0

\frac{5}{\sqrt{2}+1} + \frac{5}{\sqrt{3}+\sqrt{2}} + \frac{5}{\sqrt{4}+\sqrt{3}}

+ \dots \frac{5}{\sqrt{121}+\sqrt{120}}

Solution:

Expression :

\frac{5}{\sqrt{2}+1} + \frac{5}{\sqrt{3}+\sqrt{2}} + \frac{5}{\sqrt{4}+\sqrt{3}}

+ \dots \frac{5}{\sqrt{121}+\sqrt{120}}

Rationalizing the denominator, we get :

\left(\frac{5}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}\right) + \left(\frac{5}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-2}{\sqrt{3}-2}\right)

+ \left(\frac{5}{\sqrt{4}+\sqrt{3}} \times \frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}-\sqrt{3}}\right)

+ \dots + \left(\frac{5}{\sqrt{121}+\sqrt{120}} \times \frac{\sqrt{121}-\sqrt{120}}{\sqrt{121}-\sqrt{120}}\right)

Using,

(a - b)(a + b) = a^2 - b^2

\frac{5 \times (\sqrt{2} - 1)}{2-1} + \frac{5 \times (\sqrt{3} - \sqrt{2})}{3-2}

+ \frac{5 \times (\sqrt{4} - \sqrt{3})}{4-3}

+ \dots + \frac{5 \times (\sqrt{121} - \sqrt{120})}{121-120}

Thus, after cancelling the positive and negative terms alternatively, we are left with :

-5 + 5(\sqrt{121}) = -5 + 55 = 50

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