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Arithmetic Progression

0 for all i, ...

Question

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

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Solution

The correct option is A
As given $a_{2} - a_{1} = a_{3} - a_{2} = . . . . . = a_{n} - a_{n - 1} = d$
Where d is the common difference of the given A.P.
Also $a_{n} = a 1 = (n - 1) d .$
Then by rationalising each term,
$\frac{1}{\sqrt{a_{2}} + \sqrt{a_{1}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{2}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n}} + \sqrt{a_{n - 1}}}$
$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} . . . . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}}$

$= \frac{1}{d} {\sqrt{a_{2}} - \sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + . . . . . \sqrt{a_{n}} - \sqrt{a_{n - 1}}}$
$\frac{1}{d} {\sqrt{a_{n}} - \sqrt{a_{1}}} = \frac{1}{d} (\frac{a_{n} - a_{1}}{{\sqrt{a}}_{n} - \sqrt{a_{1}}})$
$\frac{1}{d} {\frac{(n - 1) d}{{\sqrt{a}}_{n} + {\sqrt{a}}_{1}}} = \frac{n - 1}{\sqrt{a_{n}} - \sqrt{a_{1}}}$
Trick : Put $a_{1} = a_{2} = a_{3} => d = 0$

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