<s>[INST] Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$ The given task is:
The following forms an arithmetic sequence: $$\frac{1}{b-a}, \frac{1}{2b}, \frac{1}{b-c}.$$
Show, that $a, b, c$ forms an geometric sequence.
It's easily enough to understand that $$ \frac{1}{2b}-\frac{1}{b-a}=\frac{1}{b-c}-\frac{1}{2b} \iff \frac{a+b}{a-b}=\frac{b+c}{b-c}$$ and that I need to prove that $$\frac{b}{a}=\frac{c}{b},$$ but between the two I just make it more and more complicated.
 [/INST] We have
$$
\frac1b = \frac{1}{b-a} + \frac1{b-c}\\
(b-a)(b-c) = (b-c + b-a)b\\
b^2 - ab - bc + ac = 2b^2-bc - ab\\
ac = b^2
$$
and we are done.
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