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If $\frac{1}{b-a}, \frac{1}{2b}, \frac{1}{b-c}$ are the terms of an arithmetic sequence, prove that $b^2 =ac$.

I have no idea where to even start. Any help would be appreciated.

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    $\begingroup$ $\text{second term}-\text{first term}=\text{third term}-\text{second term}$ $\endgroup$ – chubakueno Jun 2 '14 at 0:23
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    $\begingroup$ Hello, welcome to math.se here is a tutorial for mathjax $\endgroup$ – Jorge Fernández Hidalgo Jun 2 '14 at 0:23
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    $\begingroup$ @chubakueno, OP doesn't specify the terms are consecutives in the arithmetic sequence. $\endgroup$ – Fabien Jun 2 '14 at 0:26
  • $\begingroup$ @Fabien you have good point but in that case his conclusion will also be not true. $\endgroup$ – Anurag A Jun 2 '14 at 0:30
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    $\begingroup$ @Fabien I didn't labeled the individual terms either :) OP, please clarify that. It is a very reasonable assumtion, though, since otherwise we would not have enough information. $\endgroup$ – chubakueno Jun 2 '14 at 0:31
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A sequence is arithmetic if the difference between successive terms is constant. In other words to get each term of the sequence you need to add the same fixed number to the previous term. $$a_{n+1}=a_n+d$$

Let $a_1=\frac{1}{b-a}, a_2=\frac{1}{2b}$ and $a_3=\frac{1}{b-c}$. So if these terms form an arithmetic sequence then, $$a_2-a_1 =a_3-a_2.$$ Now substitute and see what happens.

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The other way is to think in terms of an arithmetic mean. In an arithmetic progression, if $a_k, a_{k+1}, a_{k+2}$ are consecutive terms, then the middle term is the arithmetic mean of the ones surrounding it, or, equivalently, $2a_{k+1} = a_k + a_{k+2}$.

Using this saves you a teensy bit of algebra. It's essentially equivalent to the other answer, but I just wanted to offer another method to think about this sort of problem.

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