Relationship between arithmetic mean, geometric mean, and harmonic mean.
## Core concept
For any set of positive numbers, three classical averages exist in a fixed order relationship. This relationship holds universally and is critical for understanding when to apply each measure.
The fundamental inequality: - Arithmetic Mean (AM) ≥ Geometric Mean (GM) ≥ Harmonic Mean (HM) - Equality holds only when all observations are identical
Why this matters: - AM is pulled up by outliers; HM is pulled down by very small values; GM balances both - Different data contexts demand different measures - Exam questions test whether you pick the *correct* average for the scenario
## Formula / rule
For observations $x_1, x_2, \ldots, x_n$ (all positive):
| Average | Formula | Use case | |---------|---------|----------| | AM | $\frac{x_1 + x_2 + \cdots + x_n}{n}$ | Heights, ages, general aggregates | | GM | $\sqrt[n]{x_1 \cdot x_2 \cdots x_n}$ | Growth rates, ratios, index numbers | | HM | $\frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}}$ | Speeds, rates, inverse quantities |
Relationship formulas: - $AM \times HM = (GM)^2$ - If AM = $a$ and HM = $h$, then $GM = \sqrt{a \times h}$
## Common exam applications
1. Speed-related problems (use HM) - Average speed over a journey requires harmonic mean of segment speeds - Average of reciprocals (1/speed) is not the same as speed average
2. Growth rate problems (use GM) - Population growth, compound annual growth rate (CAGR) - Price indices, inflation measurement
3. Verification questions - Given three averages, check if AM ≥ GM ≥ HM holds - If inequality is violated, data set is invalid
4. Combined data (AM relationship) - When combining groups, weighted means apply - Relationship still holds for each group individually
## Worked example
Question: A car travels 100 km at 50 km/h and returns 100 km at 75 km/h. What is the average speed?
Solution: - Do NOT use arithmetic mean: $(50 + 75)/2 = 62.5$ km/h ✗ - Use harmonic mean (correct for speeds):
$$HM = \frac{2}{\frac{1}{50} + \frac{1}{75}} = \frac{2}{\frac{3+2}{150}} = \frac{2 \times 150}{5} = 60 \text{ km/h}$$
Verification: Total distance = 200 km; Total time = 2 + 1.33 = 3.33 hours; Average = 200/3.33 = 60 km/h ✓
## Common mistakes
| Mistake | Why it's wrong | Correct approach | |---------|---|---| | Using AM for speeds/rates | Different units (time in denominator) change weighting | Always use HM for inverse quantities | | Forgetting all values must be positive | GM undefined for negatives; HM undefined for zero | Check data range first | | Ignoring frequencies in grouped data | Weighted versions are needed | Apply weights to each average formula | | Assuming equality in AM = GM = HM | Only true when all observations identical | Test with simple data (e.g., 2, 3) |
## Key takeaway for exams
"When in doubt: AM for totals, GM for growth, HM for rates." The inequality $AM \geq GM \geq HM$ is a safety check — if your answer violates this, recalculate.