Q.Find the geometric mean of the numbers 2, 4 and 8.

The three means satisfy $G^2=A\times H$ for two numbers.

For two positive numbers $a$ and $b$:

$A=\frac{a+b}{2},\quad G=\sqrt{ab},\quad H=\frac{2ab}{a+b}.$

Then

$A\times H=\frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab=(\sqrt{ab})^2=G^2.$ …

AM $=5\Rightarrow a+b=10$; GM $=4\Rightarrow ab=16$; solving gives the numbers $2$ and $8$.

Let the two numbers be $a$ and $b$.

Step 1 — Arithmetic Mean:

$\dfrac{a+b}{2}=5 \Rightarrow a+b=10$

Step 2 — Geometric Mean:

$\sqrt{ab}=4 \Rightarrow ab=16$

Step 3 — form the quadratic whose roots are $a$ and $b$:

…

$G=\sqrt{2\cdot 8}=4$.

For two positive values $a$ and $b$, the geometric mean is

$G=\sqrt{ab}=\sqrt{2\times 8}=\sqrt{16}=4.$

…

$\frac{1}{3}\big(\frac15+\frac17+\frac18\big)=\frac{1}{3}\cdot\frac{131}{280}=\frac{131}{840}$.

Take the common denominator $\text{LCM}(5,7,8)=280$:

$\frac15=\frac{56}{280},\quad \frac17=\frac{40}{280},\quad \frac18=\frac{35}{280}.$

Sum of reciprocals:

$\frac{56+40+35}{280}=\frac{131}{280}.$ …

$H=\dfrac{n}{\sum (1/x_i)}=\dfrac{3}{2+3+5}=0.3$.

Harmonic mean:

$H=\frac{n}{\sum \frac{1}{x_i}}.$

Here the values are $x_1=\tfrac12,\ x_2=\tfrac13,\ x_3=\tfrac15$, whose reciprocals are …

$G=(8\times125\times343)^{1/3}=2\times5\times7=70$.

For three values the geometric mean is

$G=\sqrt[3]{x_1 x_2 x_3}=\sqrt[3]{8\times125\times343}.$

Taking cube roots factor by factor: …