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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...

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    $\begingroup$ You mean that $a_1, a_2, \ldots, a_n$ are an arithmetic progression? You should also require that none of the $a_i$ is zero. $\endgroup$ Oct 3, 2014 at 10:32
  • $\begingroup$ oops yea, my bad (fixed it) $\endgroup$
    – Ashkan
    Oct 3, 2014 at 10:46
  • $\begingroup$ The tag (theorem-provers) is for questions about software designed for checking formal proofs or assisting with writing them, see the tag-wiki. It is not intended for all questions which are about proofs of theorems. $\endgroup$ Feb 9, 2015 at 14:45

1 Answer 1

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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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  • $\begingroup$ That partially makes sense to me but I'd really appreciate it if you could provide a more detailed explanation. $\endgroup$
    – Ashkan
    Oct 3, 2014 at 10:50
  • $\begingroup$ @Ashkan Which step was not clear to you? $\endgroup$
    – taninamdar
    Oct 3, 2014 at 10:50
  • $\begingroup$ Never mind, I read it again and it all made perfect sense, Thank you very much for your answer $\endgroup$
    – Ashkan
    Oct 3, 2014 at 10:58
  • $\begingroup$ beautiful answer, Thanks again $\endgroup$
    – Ashkan
    Oct 3, 2014 at 11:05
  • $\begingroup$ @Ashkan You're welcome. $\endgroup$
    – taninamdar
    Oct 3, 2014 at 11:06

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