Arithmetic Mean, Geometric Mean, Harmonic Mean and their relations

Question

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 $c$

Answer 1

$$a=\frac{b+c}2\tag{AM}$$ $$b=\sqrt{ac}\tag{GM}$$

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$$2a=b+c\tag{from AM,(1)}$$ $$2a=\sqrt{ac}+a\tag{from GM $b=\sqrt{ac}$}$$ $$a=\sqrt{ac}$$ $$\sqrt a=\sqrt c\implies a=c$$ 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: $$\frac2c=\frac1a+\frac1b\tag{since $a=b=c$}$$

Answer 2

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, solving 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.

Answer 3

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.

Answer 4

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 $~H=\dfrac{2uv}{u+v}$

Answer 5

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

Answer 6

Define $$ d_1 := 2a-b-c,\quad d_2 := b^2-ac,\quad d_3 := (a+b)c-2ab $$ where $\,a,b,c\,$ are positive real numbers. Now

• $\,d_1=0\,$ is equivalent to $\,a\,$ being the arithmetic mean of $\,b\,$ and $\,c.$
• $\,d_2=0\,$ is equivalent to $\,b\,$ being the geometric mean of $\,c\,$ and $\,a.$
• $\,d_3=0\,$ is equivalent to $\,c\,$ being the harmonic mean of $\,a\,$ and $\,b.$

Verify the simple algebraic identity

$$ bd_1 + d_2 + d_3 = b(2a-b-c) + (b^2-ac) + ((a+b)c-2ab) = 0. $$

Given both $\,d_1 = 0\,$ and $\,d_2 = 0,\,$ then the identity implies that $\,d_3 = 0.\,$ This proves the requested result

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 $b$.