Identity

$G^{2} = A \times H$ for two numbers

Statistics • Mean

Formula

G^{2} = A \times H

For two positive numbers

a

and

b

with arithmetic mean

A = \frac{a + b}{2}

, geometric mean

G = ab

and harmonic mean

H = \frac{2 ab}{a + b}

, the three satisfy

G^{2} = A \times H

— so

G

is itself the geometric mean of

A

and

H

. Any one of the three is therefore fixed by the other two, without finding

a

and

b

How to use it

1Given two of $A$ , $G$ , $H$ , read off the third in one line: $A = \frac{G ^{2}}{H}$ , $H = \frac{G ^{2}}{A}$ , $G = A H$ .
2Check the result against $H \leq G \leq A$ — an $A$ that comes out below $G$ , or an $H$ above it, means the division went the wrong way.
3To recover the pair itself, use $a + b = 2 A$ and $ab = G^{2}$ , which make $a$ and $b$ the roots of $t^{2} - 2 A t + G^{2} = 0$ .

The arithmetic mean of two positive numbers is

5

and their geometric mean is

4

. Find their harmonic mean.

Step-by-step

1
$G^{2} = A \times H \Rightarrow 4^{2} = 5 \times H$
Why: Two numbers, so the relation applies.
2
$H = \frac{16}{5} = 3.2$
Why: Consistent with $H \leq G \leq A$ , i.e. $3.2 \leq 4 \leq 5$ .

Answer

\frac{16}{5}

Watch out

The relation is special to two numbers: for three or more values the three means satisfy only the ordering

H \leq G \leq A

, and no product relation at all. Both numbers must be positive. It also says nothing about

a

and

b

separately, and a real pair exists only when

G \leq A

, since

t^{2} - 2 A t + G^{2} = 0

needs

A^{2} \geq G^{2}

Builds on

n

n

Also known as

AM GM HM relation, G squared equals A times H, AM x HM = GM^2, geometric mean is the GM of the AM and HM