If $a_{1},a_{2},\ldots a_{n}$ are in arithmetic progression, where $a_{1} 0$ for all $i$. Then, $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + ... + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}$ is equal to

Question

If $a_{1},a_{2},\ldots a_{n}$ are in arithmetic progression, where $a_{1} > 0$ for all $i$. Then, $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + ... + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Since, $a_{1},a_{2},\ldots,a_{n}$ are in AP.
Then, $a_{2} - a_{1} = a_{3} - a_{2} = ... = a_{n} - a_{n - 1} = d$
Where $d$ is common difference
Now, $\frac{1}{\sqrt{a_{2}} + \sqrt{a_{1}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{2}}} + \ldots + \frac{1}{\sqrt{a_{n}} + \sqrt{a_{n - 1}}}$
$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + \ldots + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{d}$
$= \frac{1}{d}(\sqrt{a_{n}} - \sqrt{a_{1}}) 	imes \frac{\sqrt{a_{n}} + \sqrt{a_{1}}}{\sqrt{a_{n}} + \sqrt{a_{1}}}$
$= \frac{1}{d}\left( \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}} ight) = \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

Q. If $a_{1},a_{2},\ldots a_{n}$ are in arithmetic progression, where $a_{1} > 0$ for all $i$. Then, $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + ... + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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