Explanation

(a)
Since, $a_{1},\ a_{2},\ a_{3},\ldots,a_{n}$ are in AP.
Then, $a_{2} - a_{1} = a_{3} - a_{2} = ... = a_{n} - a_{n - 1} = d$
Where $d$ is the common difference of the give AP
Also, $a_{n} = a_{1} + (n - 1)d$
Then, by rationalizing each term
$\frac{1}{\sqrt{a_{2}} + \sqrt{a_{1}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{2}}} + ... + \frac{1}{\sqrt{a_{n}} + \sqrt{a_{n - 1}}}$
$= \frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + ... + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}}$
$= \frac{1}{d}\ (\sqrt{a_{2}} - \sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + ... + \sqrt{a_{n}} - \sqrt{a_{n - 1}}$
$= \frac{1}{d}\ \left( \sqrt{a_{n}} - \sqrt{a_{1}} \right) \times \frac{\sqrt{a_{n}} + \sqrt{a_{1}}}{\sqrt{a_{n}} + \sqrt{a_{1}}}$
$= \frac{1}{d}\left( \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}} \right) = \frac{1}{d}\left( \frac{(n - 1)d}{\sqrt{a_{n}} + \sqrt{a_{1}}} \right) = \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}$