<s>[INST] Prove that $\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$ There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me : 

If $a_1,a_2,a_3,...,a_n$ IS an arithmetic progression and $a_n$ is NOT equal to 0 then prove the following statement :
$\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$

Thanks in advance...
 [/INST] Here's sketch, not complete proof.
Since $A$ is an arithmetic progression, Let $a_{i+1}-a_i = d$
$$\begin{align}\dfrac{1}{a_ia_{i+1}} &= \dfrac{d}{d}\dfrac{1}{a_ia_{i+1}}
\\&= \dfrac{1}{d}\dfrac{a_{i+1}-a_i}{a_ia_{i+1}} \qquad \text{since } a_{i+1}-a_i = d
\\&= \dfrac{1}{d}\left(\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}}\right)\end{align}$$
Therefore, $$\begin{align} \dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots + \dfrac{1}{a_{n-1}a_n}&=
\sum\limits_{i = 1}^{n-1}\dfrac{1}{a_ia_{i+1}} \\&= \dfrac{1}{d}\sum\limits_{i = 1}^{n-1}\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}} \\&= \dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac{1}{a_2}+\dfrac{1}{a_2}-\cdots - \dfrac{1}{a_{n-1}} + \dfrac{1}{a_{n-1}}-\dfrac{1}{a_n} \right)\\&=\dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac1{a_n}\right)\end{align}$$
Now, take LCM and use the fact that $a_n = a_1 +(n-1)d$
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