CAT 2019

Slot 1

QA

Question & Solution

We have, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

Now, $\frac{1}{\sqrt{a_1} + \sqrt{a_2}}$ = $\frac{\sqrt{a_2} - \sqrt{a_1}}{(\sqrt{a_2} + \sqrt{a_1})(\sqrt{a_2} - \sqrt{a_1})}$   (Multiplying numerator and denominator by $\sqrt{a_2} - \sqrt{a_1}$)

= $\frac{\sqrt{a_2} - \sqrt{a_1}}{({a_2} - {a_1}}$

=$\frac{\sqrt{a_2} - \sqrt{a_1}}{d}$   (where d is the common difference)

Similarly, $\frac{1}{\sqrt{a_2} + \sqrt{a_3}}$ = $\frac{\sqrt{a_3} - \sqrt{a_2}}{d}$ and so on.

Then the expression $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$

can be written as $\ \frac{\ 1}{d}(\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_3}+..........................\sqrt{a_{n+1}} - \sqrt{a_{n}}$

= $\ \frac{\ n}{nd}(\sqrt{a_{n+1}}-\sqrt{a_1})$ (Multiplying both numerator and denominator by n)

= $\ \frac{n(\sqrt{a_{n+1}}-\sqrt{a_1})}{{a_{n+1}} - {a_1}}$     $(a_{n+1} - {a_1} =nd)$

= $\frac{n}{\sqrt{a_1} + \sqrt{a_{n + 1}}}$