Q.For the numbers $4$ and $9$, find the Arithmetic Mean (A), Geometric Mean (G) and Harmonic Mean (H), and verify that $G^2 = A \times H$ and $A \ge G \ge H$.

Apply $A = \frac{a+b}{2}$, $G = \sqrt{ab}$, $H = \frac{2ab}{a+b}$, then check the two relations.

With $a = 4$ and $b = 9$:

• Arithmetic Mean: $A = \dfrac{a + b}{2} = \dfrac{4 + 9}{2} = \dfrac{13}{2} = 6.5$.
• Geometric Mean: $G = \sqrt{ab} = \sqrt{4 \times 9} = \sqrt{36} = 6$.
• Harmonic Mean: $H = \dfrac{2ab}{a + b} = \dfrac{2(4)(9)}{4 + 9} = \dfrac{72}{13} \approx 5.538$.

Verify $G^2 = A \times H$: $G^2 = 6^2 = 36$; and $A \times H = 6.5 \times \dfrac{72}{13} = \dfrac{6.5 \times 72}{13} = \dfrac{468}{13} = 36$. Equal ✓. …

The relation $G^2 = A\cdot H$ gives an independent way to find any one mean from the other two — e.g. once $A = 6.5$ and $G = 6$ are known, $H = \dfrac{G^2}{A} = \dfrac{36} …

Computing HM as $\frac{a+b}{2ab}$ (the reciprocal of the correct formula) — that is actually $\frac1H$. The HM itself is $\frac{2ab}{a+b}$. A quick sanity check: HM must be the smalles …