algebra precalculus - Proving an identity involving terms in arithmetic progression.

If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:

$(1)\large \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$

$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n} \biggr)$

Answer

This question is now old enough for some more complete answers.

For number 1:

$$\sum_{k=1}^{n-1} \frac{1}{ \sqrt{a_k}+ \sqrt{a_{k+1}}} = \sum_{k=1}^{n-1} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$
where $d$ is the common difference,
$$= \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right) = \frac{a_n - a_1}{d(\sqrt{a_n} + \sqrt{a_1})}$$
$$= \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$

And for number 2:

$$S = 2 \sum_{k=1}^n \frac{1}{a_k} = \left( \frac{1}{a_1} + \frac{1}{a_n} \right) + \left( \frac{1}{a_2} + \frac{1}{a_{n-1}} \right) + \cdots + \left( \frac{1}{a_n} + \frac{1}{a_1} \right)$$

$$= \sum_{k=1}^n \frac{a_{n-k+1}+ a_k}{a_k a_{n-k+1}}.$$

Now $a_{n-k+1}+ a_k = 2a_1 + (n-1)d = a_1 + a_n$ and so

$$S = (a_1+a_n) \sum_{k=1}^n \frac{1}{a_k a_{n-k+1}}$$

from which the result follows.