Question 68

If $$a_1, a_2, ......$$ are in A.P., then, $$\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + ....... + \frac{1}{\sqrt{a_n} + \sqrt{a_{n + 1}}}$$ is equal to

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)

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

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

Video Solution

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