Arithmetic Mean, Geometric Mean, Harmonic Mean and their relations 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 $c$
 A: 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. 
A: Hint: It is a rather well-known and beautiful fact, that the geometric mean of two numbers is also the geometric mean between the arithmetic and harmonic mean of those two numbers: 
$\qquad\qquad\quad G=\sqrt{uv}\iff G=\sqrt{AH},~$ where $~A=\dfrac{u+v}2~$ and $~H=\dfrac{2uv}{u+v}$
A: a is the arithmetic mean between b and c 
       a = (b+c)/2 --------------(1)

‘b’ is the geometric mean between a and c
b = √ac  OR b^2= ac - - - - - - - - (2)

by Multiplying  eqn(1) by ‘b’ and replacing b2 by  eqn (2) in (1) we get,
ab = b ((b+c)/2)
ab =  (b2 +bc )/2
ab = (ac +bc )/2
ab = c(a+b) )/2
c = 2ab /(a+b)
hence c is the haromonic mean of a and b
A: $$a=\frac{b+c}2\tag{AM}$$
$$b=\sqrt{ac}\tag{GM}$$

$$2a=b+c\tag{from AM,(1)}$$
$$2a=\sqrt{ac}+a\tag{from GM $b=\sqrt{ac}$}$$
$$a=\sqrt{ac}$$
$$\sqrt a=\sqrt c\implies a=c$$
Put $a=c$ in (1)
$$2a=b+a\implies b=a$$
So $a=b=c$, so c is the harmonic mean between a and b as:
$$\frac2c=\frac1a+\frac1b\tag{since $a=b=c$}$$
A: I think $a=b=c$ (and therefore all the means are the same).
Given the two equations, you get
$$
ac = b^2 = (2a-c)^2 = 4a^2 - 4ac +c^2
$$
Hence
$$
c^2 -5ac +4a^2 = 0
$$
which, solvinc for $c$, yields $c=a$ or $c=4a$. Now, since $b = 2a-c$, we can only have $b=a$ or $b=-2a$. Since $b$ cannot be negative (otherwise your second assumption would not hold), we get $b=a$ too.
