[Math] Prove consecutive terms in a geometric sequence and consecutive terms in an arithmetic sequence.

Question

Let $a,b$ and $c$ be consecutive numbers in a geometric sequence, where $a+b\neq 0$ and $b+c\neq 0$.
Show that $$ \frac{2ab}{a+b}, b, \frac{2bc}{b+c} $$ are consecutive terms in an arithmetic sequence.

Answer 1

Since $a$, $b$, $c$ are consecutive numbers in a geometric sequence, there exist a real number $r$ such that $b=ar$ and $c=ar^2$. Since $a+b\neq0$ and $b+c\neq0$, it follows $a\neq0$, $r\neq-1$, and $r\neq 0$; then observe \begin{align} b-\dfrac{2ab}{a+b}&=\frac{ab+b^2-2ab}{a+b}=\frac{b^2-ab}{a+b}=\frac{b(b-a)}{a+b}=\frac{ar(ar-a)}{a+ar}=\frac{ar(r-1)}{1+r} \end{align} On the other hand, $$\dfrac{2bc}{b+c}-b=\frac{2bc-b^2-bc}{b+c}=\frac{bc-b^2}{b+c}=\frac{b(c-b)}{b+c}=\frac{ar(ar^2-ar)}{ar+ar^2}=\frac{ar(r-1)}{1+r}$$ From this the result follows.