If $a$ be the arithmetic mean between $b$ and $c$, $b$ be the geometric mean between $c$ and $a$ then prove that $c$ is the harmonic mean between $a$ and $b$. I expressed $a$ as $$a=\frac{(b+c)}{2}$$ $$b=\sqrt {ac}$$ . I solved the equations but I could not evaluate for $c$
[Math] Arithmetic Mean, Geometric Mean, Harmonic Mean and their relations
algebra-precalculusmeanssystems of equations
Best Answer
We know that $c=2a-b=b^2/a$ ($a$ and $c$ must have the same sign). We have to prove that $$ c=\frac{2}{(1/a+1/b)}=\frac{2ab}{a+b} $$ So let's take the difference: $$ 2a-b-\frac{2ab}{a+b}=\frac{2a^2+2ab-ab-b^2-2ab}{a+b}=\frac{a(2a-b)-b^2}{a+b} $$ Since $b^2=a(2a-b)$ this difference is zero.