Harmonic progression — definition, HM: Visual walkthrough

We will only assume you can:

put numbers on a line and measure gaps between them, and
flip a number to its reciprocal (turn $4$ into $\frac{1}{4}$ ).

Everything else we build.

Step 1 — Three numbers, one question

WHAT. We start with two positive numbers. Call the smaller one $a$ and the larger one $b$ . We want to slip a third number between them — call it $H$ — chosen by a special "harmonic" rule.

WHY. The whole idea of a mean is: given two ends, find one middle number that fairly represents them. There are several fair rules (we compare them later). The harmonic rule is the one that matters when quantities combine as reciprocals — like times, resistances, or lens powers.

PICTURE. Look at the number line below. The two ends $a = 4$ and $b = 6$ are marked in chalk-blue. The mystery point $H$ (pink) sits somewhere between them — we don't know where yet. That's the hole we're going to fill.

Figure — Harmonic progression — definition, HM

Step 2 — The trick: flip everything to reciprocal-land

WHAT. Replace every number by its reciprocal: $a \to \frac{1}{a}$ , $H \to \frac{1}{H}$ , $b \to \frac{1}{b}$ . The reciprocal of a number is just $1$ divided by it — it measures "how many fit into one".

WHY. In the parent note, an HP is defined as a list whose reciprocals form an ordinary evenly-spaced list (an AP). Evenly-spaced numbers are the easiest thing in mathematics to reason about — equal gaps. So instead of fighting $a, H, b$ directly, we jump to the mirror world where the rule becomes "equal gaps".

PICTURE. Below, the top line is our original world ( $a, H, b$ , increasing left to right). The bottom line is reciprocal-land. Notice something important: flipping reverses the order — the biggest number $b$ becomes the smallest reciprocal $\frac{1}{b}$ . Follow the dashed arrows down from each point.

Step 3 — The harmonic rule made visible: equal gaps

WHAT. We now demand the harmonic condition: in reciprocal-land the three points $\frac{1}{a}, \frac{1}{H}, \frac{1}{b}$ must be evenly spaced. That is, the gap from $\frac{1}{a}$ to $\frac{1}{H}$ equals the gap from $\frac{1}{H}$ to $\frac{1}{b}$ .

WHY. This is exactly what " $a, H, b$ in HP" means (parent definition: reciprocals form an AP). "Evenly spaced" is the AP condition written as a picture. This single geometric demand is the entire content of the harmonic mean.

PICTURE. The two chalk-yellow braces below are the two gaps. The harmonic rule says: make these two braces the same length. Written with symbols right on the picture:

$left gap \frac{1}{H} - \frac{1}{a} = right gap \frac{1}{b} - \frac{1}{H}$

$\frac{1}{H} - \frac{1}{a}$ ::: how far $\frac{1}{H}$ sits to the right of $\frac{1}{a}$ (the left brace).
$\frac{1}{b} - \frac{1}{H}$ ::: how far $\frac{1}{b}$ sits to the right of $\frac{1}{H}$ (the right brace).

Step 4 — Solve the equal-gap equation for $\frac{1}{H}$

WHAT. Take the equation from Step 3 and gather all the $\frac{1}{H}$ terms on one side.

WHY. The picture says "left brace = right brace". Turning that picture into algebra and isolating $\frac{1}{H}$ tells us exactly where the middle point lands.

Start: $\frac{1}{H} - \frac{1}{a} = \frac{1}{b} - \frac{1}{H} .$ Move both $\frac{1}{H}$ to the left, both ends to the right: $\frac{1}{H} + \frac{1}{H} = \frac{1}{a} + \frac{1}{b} .$ The left side is just two copies of $\frac{1}{H}$ , i.e. $2 \cdot \frac{1}{H}$ : $\frac{2}{H} = \frac{1}{a} + \frac{1}{b} .$ Divide both sides by $2$ : $\frac{1}{H} = \frac{1}{2} (\frac{1}{a} + \frac{1}{b})$

$\frac{1}{H}$ ::: the reciprocal of the middle number.
$\frac{1}{2} (\frac{1}{a} + \frac{1}{b})$ ::: the plain average of the two end reciprocals.

PICTURE. The midpoint of two evenly-spaced ends is literally their average — the middle brace lands halfway. The figure marks $\frac{1}{H}$ dead-centre between $\frac{1}{b}$ and $\frac{1}{a}$ .

Step 5 — Tidy the right-hand side into one fraction

WHAT. Combine $\frac{1}{a} + \frac{1}{b}$ over a common denominator before flipping.

WHY. To flip a fraction cleanly, we want $\frac{1}{H}$ written as a single fraction, not a sum. Then flipping is just turning it upside-down.

$\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a + b}{ab} .$

$\frac{b}{ab}$ ::: $\frac{1}{a}$ rewritten with denominator $ab$ (multiply top and bottom by $b$ ).
$\frac{a}{ab}$ ::: $\frac{1}{b}$ rewritten with denominator $ab$ .
$\frac{a + b}{ab}$ ::: their sum — numerator adds, denominator stays $ab$ .

Feed that into Step 4's box: $\frac{1}{H} = \frac{1}{2} \cdot \frac{a + b}{ab} = \frac{a + b}{2 ab} .$

PICTURE. The figure stacks the fractions as chalk bars so you see $\frac{b}{ab}$ and $\frac{a}{ab}$ combine into one bar $\frac{a + b}{ab}$ .

Step 6 — Flip back to the real world

WHAT. We have $\frac{1}{H}$ . The number we actually wanted is $H$ , so turn the fraction upside-down.

WHY. In Step 2 we entered reciprocal-land by flipping. To read the answer as a real position on our original line, we must leave reciprocal-land the same way — by flipping again. "Flip → solve → flip back" is the entire strategy.

$\frac{1}{H} = \frac{a + b}{2 ab} ⟹ H = \frac{2 ab}{a + b} .$

$\frac{2 ab}{a + b}$ ::: the harmonic mean. Read it as "Two · $a$ · $b$ over their Sum" — the mnemonic TABS.

PICTURE. Both worlds shown together: the pink point in reciprocal-land was the midpoint; flipping the whole line lands it at $H = 4.8$ back in the real world — closer to the smaller end $4$ than to $6$ . That leftward pull is a permanent feature of HM (proved in Step 8).

Step 7 — The degenerate case: what if $a = b$ ?

WHAT. Suppose the two ends are the same number, $a = b$ . What does the formula do?

WHY. A good mean must return that shared value — the average of $5$ and $5$ has to be $5$ . We must check the formula doesn't misbehave or divide by zero here.

$H = \frac{2 \cdot a \cdot a}{a + a} = \frac{2 a ^{2}}{2 a} = a . ✅$

PICTURE. All three points collapse onto one spot; both gaps shrink to zero (equal, trivially). The denominator $a + b = 2 a$ is safely non-zero as long as $a \neq = 0$ .

Step 8 — Why $H$ always leans toward the smaller number (AM vs HM)

WHAT. Compare $H$ with the ordinary average $AM = \frac{a + b}{2}$ . For $a = 4, b = 6$ : AM $= 5$ , HM $= 4.8$ . The HM is smaller.

WHY. This isn't a coincidence — it's the whole reason HM exists as a separate tool. Averaging in reciprocal-land squashes big numbers (because big numbers have tiny reciprocals), so the answer is dragged toward the smaller end. That's exactly what you want for average speed over equal distances: the slow leg eats more time, so it should dominate.

PICTURE. One line with three middles marked: HM (pink, $4.8$ ), GM $= ab \approx 4.90$ (yellow), AM (blue, $5$ ). They always sit in the order $AM \geq GM \geq HM,$ with equality only when $a = b$ (Step 7). And they lock together through $GM^{2} = AM \cdot HM$ .

The one-picture summary

Everything above on a single board: enter reciprocal-land (flip), make the gaps equal (AP condition), take the midpoint (average of reciprocals), flip back. Out drops $H = \frac{2 ab}{a + b}$ , sitting left-of-centre, below AM and GM.

Recall Feynman retelling — the whole walkthrough in plain words

We had two numbers and wanted a fair middle. But "fair" here means fair to reciprocal quantities like times or resistances. So we flipped both numbers into reciprocal-land, where "fair middle" simply means "sit exactly halfway" — equal gaps on each side. Halfway between two numbers is just their average, so the reciprocal of our answer is the average of the two reciprocals. Then we flipped that answer back to the real world. Cleaning up the fractions gave Two-a-b-over-Sum, the harmonic mean $\frac{2 ab}{a + b}$ . We checked it behaves: it gives back $a$ when both ends are equal, it never divides by zero for positive numbers, and it always leans toward the smaller number — sitting below the geometric and arithmetic means, all three locked by $GM^{2} = AM \cdot HM$ .

Connections

Parent: HP & HM — the result this page derives in pictures.
Arithmetic Progression — the "equal gaps" world we jump into (Step 2–3).
Arithmetic Mean — the midpoint rule we borrow; compared in Step 8.
Geometric Mean / Geometric Progression — the middle mean in the AM–GM–HM lock.
Average speed and rates — why leaning-to-the-slow-side (Step 8) is physically correct.
Resistors in parallel / Lens formula — reciprocals that add, the native home of HM.