<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] Taking it from where you left off, use cross-products and simplify
$$(a+b)(b-c) = (a-b)(b+c) \iff \color{blue}{ab}-ac+b^2\color{green}{-bc} = \color{blue}{ab}+ac-b^2\color{green}{-bc}$$
$$2b^2 = 2ac \iff b^2 = ac \iff \frac{b}{a} = \frac{c}{b}$$
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