<s>[INST] 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}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots +  \frac{1}{a_n} \biggr)$

 [/INST] 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.
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