Prove a geometric sequence a, b, c from the arithmetic progression $1/(b-a)$, $1/2b$, $1/(b-c)$

Question

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.

Answer 1

Taking it from where you left off, use cross-products and simplify

$$(a+b)(b-c) = (a-b)(b+c) \iff \color{blue}{ab}-ac+b^2\color{green}{-bc} = \color{blue}{ab}+ac-b^2\color{green}{-bc}$$

$$2b^2 = 2ac \iff b^2 = ac \iff \frac{b}{a} = \frac{c}{b}$$

Answer 2

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.