Harmonic progression — definition, HM: Worked examples

This page is the "no surprises" drill for Harmonic Progression (the wiki-link points to a companion note; if you are reading this standalone, everything you need is defined right below). We march through every kind of case an HP/HM problem can throw at you — friendly positives, sign traps, a genuine sign-changing HP, the forbidden zero, the degenerate "all equal" case, a limit, a real-world word problem, and an exam-style twist. Each example first asks you to forecast the answer (guess before you compute — that is where learning sticks), then walks every step with a why, then verifies by plugging back.

Definition Self-contained vocabulary (nothing assumed)

An Arithmetic Progression (AP) is a list where each term differs from the previous by the same fixed amount, called the common difference $D$ . Example: $2, 5, 8, 11$ has $D = 3$ . We write the first term as $A$ , and the $n$ -th term is $A + (n - 1) D$ (start at $A$ , then add $D$ a total of $n - 1$ times).
A Harmonic Progression (HP) is a list whose reciprocals (the "flipped" numbers $\frac{1}{a _{1}}, \frac{1}{a _{2}}, \dots$ ) form an AP. So $6, 3, 2$ is an HP because $\frac{1}{6}, \frac{1}{3}, \frac{1}{2}$ is an AP.
The Harmonic Mean (HM) of two numbers $a, b$ is the single number $H$ that makes $a, H, b$ an HP; it works out to $H = \frac{2 ab}{a + b}$ .
The Arithmetic Mean (AM) of $a, b$ is the ordinary average $\frac{a + b}{2}$ ; the Geometric Mean (GM) is $ab$ .
The symbol $\frac{1}{0}$ means "one divided by zero." Division by zero has no value — it is undefined. So no HP term may ever be $0$ , because flipping it would demand $\frac{1}{0}$ .

The scenario matrix

Here is the full landscape. Every example below is tagged with the cell it covers, and together they touch every row.

Cell	Case class	What makes it tricky	Covered by
A	All-positive HP, find a term	plain bridge use	Ex 1
B	HM of two positives	flip-back trap	Ex 2
C	Sign-changing HP (mixed signs across the list)	reciprocal-AP crosses $0$ ; must avoid the undefined $\frac{1}{0}$	Ex 3
D	Zero / degenerate input	reciprocal of $0$ is illegal; equal terms $\Rightarrow D = 0$	Ex 4
E	Limiting behaviour (term $\to \infty$ , HM as $b \to \infty$ )	denominators shrinking to $0$	Ex 5
F	Real-world word problem (equal-distance speed)	know why HM, not AM	Ex 6
G	Insert $k$ harmonic means between two numbers	build the reciprocal-AP fully	Ex 7
H	Exam twist: AM·HM = GM² used backwards	mix all three means	Ex 8

Worked example

Ex 1 — Cell A: all-positive HP, find a term The numbers $12, 6, 4, \dots$ are in HP. Find the $6$ th term.

Forecast: the terms are shrinking fast ( $12, 6, 4$ ). Guess: will the 6th term be near $2$ , or below $1.5$ ? Jot a guess.

Step 1 — Flip to reciprocals. $\frac{1}{12}, \frac{1}{6}, \frac{1}{4}$ . Why this step? HP is only defined through reciprocals — we cannot use AP tools until we cross the bridge.

Step 2 — Confirm it is an AP and read off $A, D$ . Common denominator $12$ : $\frac{1}{12}, \frac{2}{12}, \frac{3}{12}$ . Difference $= \frac{1}{12}$ each step. So $A = \frac{1}{12}$ , $D = \frac{1}{12}$ . Why this step? Without a constant difference, there is no AP and the whole method collapses.

Step 3 — 6th reciprocal via AP nth term. $A + (6 - 1) D = \frac{1}{12} + 5 \cdot \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$ . Why this step? The AP formula gives the reciprocal of the answer — never the answer itself.

Step 4 — Flip back. $a_{6} = \frac{1}{1/2} = 2$ . Why this step? We are working inside HP-land; the final term must be un-flipped.

Verify: reciprocals should read $\frac{1}{12}, \frac{2}{12}, \frac{3}{12}, \frac{4}{12}, \frac{5}{12}, \frac{6}{12}$ — the 6th is $\frac{6}{12} = \frac{1}{2}$ , flip gives $2$ . Matches. (Your forecast of "near $2$ "? ✅)

The figure below plots both lists side by side: the cyan reciprocals lie on a perfectly straight line (that is the AP), while the amber HP terms bow downward toward $0$ . Trace the cyan line up by one step of $\frac{1}{12}$ from index $5$ to $6$ , then flip that height to read off $a_{6} = 2$ on the amber curve.

Figure — Harmonic progression — definition, HM

Worked example

Ex 2 — Cell B: HM of two positives (flip-back trap) Find the harmonic mean of $3$ and $5$ .

Forecast: HM is always smaller than the ordinary average $\frac{3 + 5}{2} = 4$ . So guess something below $4$ — around $3.7$ ?

Step 1 — Average the reciprocals. $\frac{1}{H} = \frac{1}{2} (\frac{1}{3} + \frac{1}{5}) = \frac{1}{2} \cdot \frac{5 + 3}{15} = \frac{1}{2} \cdot \frac{8}{15} = \frac{4}{15}$ . Why this step? Because $3, H, 5$ in HP means $\frac{1}{3}, \frac{1}{H}, \frac{1}{5}$ in AP, and an AP's middle term is the average of its neighbours.

Step 2 — Flip back (the trap!). $H = \frac{15}{4} = 3.75$ . Why this step? $\frac{4}{15}$ is $\frac{1}{H}$ , not $H$ . Forgetting to flip is the classic error from the parent note.

Step 3 — Cross-check with the direct formula. $H = \frac{2 ab}{a + b} = \frac{2 \cdot 3 \cdot 5}{3 + 5} = \frac{30}{8} = \frac{15}{4} = 3.75$ . Same. Why this step? Two independent routes to one number is the safest sanity check.

Verify: is $3, 3.75, 5$ genuinely an HP? Reciprocals $0.3333, 0.2667, 0.2$ ; differences $- 0.0667$ and $- 0.0667$ — equal. ✅ And $3.75 < 4$ as forecast.

Worked example

Ex 3 — Cell C: a sign-changing HP (mixed signs) Are $- 6, - 12, 12, 4, \dots$ in HP? If so, find the next term. (Notice the list itself switches from negative to positive — this is the genuine mixed-sign case.)

Forecast: flipping mixed signs gives mixed-sign reciprocals. For an HP those flipped numbers must climb in equal steps. Guess before checking: do the four given numbers really flip into an evenly spaced list?

Step 1 — Flip the four given numbers. Reciprocals of $- 6, - 12, 12, 4$ are $- \frac{1}{6}, - \frac{1}{12}, \frac{1}{12}, \frac{1}{4} .$ Why this step? The bridge is sign-blind; a reciprocal keeps the sign of the original number, so mixed-sign inputs give mixed-sign reciprocals. We must test these, not guess.

Step 2 — Put them over a common denominator and test the differences. In twelfths: $- 2, - 1, + 1, + 3$ (each over $12$ ). Differences: $(- 1) - (- 2) = + 1$ , then $(+ 1) - (- 1) = + 2$ , then $(+ 3) - (+ 1) = + 2$ . The steps are $+ \frac{1}{12}, + \frac{2}{12}, + \frac{2}{12}$ — not constant. Why this step? HP requires the reciprocal list to be an AP, i.e. one fixed step. Here the steps disagree. Answer to the posed question: No — $- 6, - 12, 12, 4, \dots$ is NOT in HP, so there is no "next term" to find.

Step 3 — Repair it to the nearest genuine sign-changing HP. Keep the first two reciprocals $- \frac{2}{12}, - \frac{1}{12}$ (i.e. keep $- 6, - 12$ and their step of $+ \frac{1}{12}$ ). Continuing that one fixed step forces the reciprocal-AP $- \frac{2}{12}, - \frac{1}{12}, step lands on 0 \frac{0}{12}, \frac{1}{12}, \frac{2}{12}, \dots$ Why this step? Because $- 6, - 12$ already fix a step of $+ \frac{1}{12}$ , honest continuation must pass through $0$ . That is the heart of every sign-changing HP: to go from negative reciprocals to positive ones, the AP has to cross $0$ .

Step 4 — Handle the crossover honestly. The slot at $\frac{0}{12} = 0$ would need HP term $\frac{1}{0}$ , which is undefined (one divided by zero has no value). So that position is a legal hole — the HP simply has no term there and jumps over it. Why this step? This is exactly why the original list failed: you cannot smuggle a defined term across the zero crossing; a true sign-changing HP always skips one slot.

Step 5 — Read the repaired sign-changing HP. Flip the defined reciprocals $- \frac{2}{12}, - \frac{1}{12}, \frac{1}{12}, \frac{2}{12}, \frac{3}{12}$ (simplify: $- \frac{1}{6}, - \frac{1}{12}, \frac{1}{12}, \frac{1}{6}, \frac{1}{4}$ ) back to terms: $- 6, - 12, no term (hole), 12, 6, 4.$ So a valid sign-changing HP that begins with the given $- 6, - 12$ reads $- 6, - 12, (hole), 12, 6, 4, \dots$ , and the term after $4$ comes from the next reciprocal $\frac{4}{12}$ : flip to get $\frac{1}{4/12} = 3$ . Why this step? It shows the correct object the problem was gesturing at: the given list was almost this HP but had the wrong third and fourth entries; the real continuation past $4$ is $3$ .

Verify: repaired reciprocal-AP in twelfths is $- 2, - 1, (0 hole), 1, 2, 3, 4$ — a constant step of $+ 1$ (i.e. $+ \frac{1}{12}$ ) everywhere except the skipped hole. Flipping the defined ones gives $- 6, - 12, 12, 6, 4, 3$ , a genuine HP that changes sign exactly as Cell C promises. The original list $- 6, - 12, 12, 4$ differs at the third/fourth entries, confirming it was not HP. ✅

The figure makes the crossing visible: the cyan reciprocal-AP is a straight line climbing by one fixed step and slicing through the horizontal $0$ line (the amber dashed line). Exactly where it hits $0$ there is an open circle — the hole — because flipping $0$ demands the undefined $\frac{1}{0}$ . Everywhere else, flip a cyan height to get an amber HP term.

Worked example

Ex 4 — Cell D: zero & degenerate inputs (a) Can $4, 0, - 4$ be in HP? (b) What is the HM of $7$ and $7$ ?

Forecast: (a) something smells wrong about the $0$ . (b) two equal numbers — guess the HM is just $7$ .

Part (a): Step 1 — Try to flip. The middle term is $0$ , and $\frac{1}{0}$ is undefined (one divided by zero has no value). Why this step? The very definition of HP forbids a $0$ term — you cannot take its reciprocal, so the bridge cannot even be built. Conclusion: $4, 0, - 4$ is not an HP (and no sequence containing $0$ ever is).

Part (b) — the degenerate equal case: Step 1 — Reciprocal-AP. $\frac{1}{7}, \frac{1}{H}, \frac{1}{7}$ . For these three to be an AP the common difference must be $0$ , so $\frac{1}{H} = \frac{1}{7}$ . Why this step? Equal end terms force $D = 0$ ; the middle must equal them. Step 2 — Flip back. $H = 7$ . Why this step? $\frac{1}{H} = \frac{1}{7}$ is the reciprocal of the answer; un-flipping returns the HM itself. Cross-check formula: $H = \frac{2 \cdot 7 \cdot 7}{7 + 7} = \frac{98}{14} = 7$ . Why this step? Confirms the degenerate case matches the general formula, and here AM = GM = HM = 7 — equality holds iff the numbers are equal (the boundary case of the AM–GM–HM chain).

Verify: $H = 7$ ; and $7 = 7 = 7$ collapses the whole AM–GM–HM chain to a single value. ✅

Worked example

Ex 5 — Cell E: limiting behaviour (a) In the HP $6, 3, 2, \dots$ (reciprocal-AP with $A = D = \frac{1}{6}$ ), what happens to $a_{n}$ as $n$ grows? (b) What is $HM (4, b)$ as $b \to \infty$ ?

Forecast: (a) terms keep shrinking — guess they crawl toward $0$ but never reach it. (b) with one number huge, guess the HM is pulled toward the small one.

Part (a): Step 1 — nth reciprocal. $\frac{1}{a _{n}} = \frac{1}{6} + (n - 1) \frac{1}{6} = \frac{n}{6}$ . Why this step? We need a clean expression in $n$ to take a limit. Step 2 — Flip. $a_{n} = \frac{6}{n}$ . Why this step? $\frac{n}{6}$ is the reciprocal of the HP term, so we un-flip to recover $a_{n}$ itself before studying its size. Step 3 — Let $n \to \infty$ . $\frac{6}{n} \to 0^{+}$ . Why this step? As the reciprocal-AP grows without bound, its reciprocal shrinks to $0$ — but stays positive, never touching $0$ . Conclusion: the HP terms approach $0$ from above; $0$ is a limit the HP never attains (consistent with "no term may be $0$ ").

Part (b): Step 1 — Write the HM. $H = \frac{2 \cdot 4 \cdot b}{4 + b} = \frac{8 b}{4 + b}$ . Why this step? We instantiate the two-number HM formula with $a = 4$ so the behaviour depends only on $b$ . Step 2 — Divide top and bottom by $b$ . $H = \frac{8}{\frac{4}{b} + 1}$ . Why this step? This isolates the vanishing term $\frac{4}{b}$ so the limit is obvious. Step 3 — Let $b \to \infty$ . $\frac{4}{b} \to 0$ , so $H \to \frac{8}{1} = 8$ . Why this step? Confirms the HM is bounded by twice the smaller number: a huge partner can never drag HM above $2 \times 4 = 8$ .

Verify: at $b = 1000$ , $H = \frac{8000}{1004} \approx 7.968$ — creeping up to $8$ , never exceeding it. ✅ Both forecasts hold.

The figure shows both limits at once. Left panel: the cyan HP terms $6/ n$ step downward and flatten onto the amber dashed line at $0$ without ever touching it. Right panel: the cyan curve $HM (4, b)$ rises but is pinned below the amber dashed ceiling at $8$ — visual proof that a giant partner cannot lift the HM past twice the smaller number.

Worked example

Ex 6 — Cell F: real-world word problem (equal distances) A drone flies $30 km$ out at $60 km/h$ and the same $30 km$ back at $40 km/h$ . What is its average speed for the round trip?

Forecast: the naive average is $\frac{60 + 40}{2} = 50$ . But it spends longer at the slower speed, so the true average must be below $50$ . Guess $48$ .

Step 1 — Recognise equal distances. Both legs are $30 km$ . Why this step? Equal distances (not equal times) is exactly the condition that makes HM, not AM, the right average — see Average speed and rates.

Step 2 — Total distance / total time. Time out $= \frac{30}{60} = 0.5 h$ , time back $= \frac{30}{40} = 0.75 h$ . Total distance $= 60 km$ , total time $= 1.25 h$ . $\overset{s}{ˉ} = \frac{60}{1.25} = 48 km/h .$ Why this step? Average speed is always total distance over total time — never an average of the speeds.

Step 3 — Match to the HM formula. $HM (60, 40) = \frac{2 \cdot 60 \cdot 40}{60 + 40} = \frac{4800}{100} = 48$ . Identical. Why this step? Shows the physics answer is the harmonic mean, cementing why HM exists.

Verify: units are $km / h$ ✅; $48 < 50$ as forecast ✅; both methods give $48$ . ✅

The bar figure makes the asymmetry obvious: the amber "back" bar is longer than the cyan "out" bar because the slower leg takes more time. Since the drone lingers at $40 km/h$ , the round-trip average is dragged below the midpoint $50$ to exactly the harmonic mean $48$ .

Worked example

Ex 7 — Cell G: insert harmonic means between two numbers Insert three harmonic means between $6$ and $\frac{6}{5}$ , i.e. build $6,_,_,_, \frac{6}{5}$ as an HP.

Forecast: the terms should fall smoothly from $6$ down to $1.2$ . Guess the inserted ones are roughly $3, 2, 1.5$ .

Step 1 — Flip both ends. $\frac{1}{6}$ and $\frac{5}{6}$ become the first and fifth terms of a reciprocal-AP (three inserts $\Rightarrow$ five terms total). Why this step? Inserting HMs into an HP = inserting AMs into the reciprocal-AP.

Step 2 — Find the common difference $D$ . From $A_{1} = \frac{1}{6}$ to $A_{5} = \frac{5}{6}$ in $4$ steps: $D = \frac{\frac{5}{6} - \frac{1}{6}}{4} = \frac{4/6}{4} = \frac{1}{6}$ . Why this step? $A_{5} = A_{1} + 4 D$ ; solving for $D$ spaces the AP evenly.

Step 3 — List the reciprocal-AP. $\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$ . Why this step? Each middle term is $A_{1}$ plus a multiple of $D$ .

Step 4 — Flip the three inserts back. $\frac{1}{2/6} = 3, \frac{1}{3/6} = 2, \frac{1}{4/6} = 1.5$ . Why this step? HMs live in HP-land, so each reciprocal is un-flipped. The three harmonic means are $3, 2, 1.5$ .

Verify: the full HP is $6, 3, 2, 1.5, 1.2$ ; reciprocals $\frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$ step by $\frac{1}{6}$ . ✅ Forecast nailed.

The figure lays the two lists on the same axes: the cyan reciprocal-AP is a straight, evenly spaced staircase (each rung $+ \frac{1}{6}$ ), and flipping the three interior rungs gives the amber inserted harmonic means $3, 2, 1.5$ — proof that evenly spacing the reciprocals is exactly what "inserting harmonic means" means.

Worked example

Ex 8 — Cell H: exam twist ( $GM^{2} = AM \cdot HM$ backwards) For two positive numbers, the arithmetic mean is $10$ and the geometric mean is $8$ . Find the harmonic mean, then recover the two numbers.

Forecast: since $AM \geq GM \geq HM$ , the HM sits below $8$ . Guess around $6$ .

Step 1 — Use the linking identity. $GM^{2} = AM \cdot HM \Rightarrow HM = \frac{GM ^{2}}{AM} = \frac{8 ^{2}}{10} = \frac{64}{10} = 6.4$ . Why this step? This identity (linking Geometric Mean to Arithmetic Mean) lets us get HM without ever knowing the numbers.

Step 2 — Recover $a + b$ and $ab$ . $AM = 10 \Rightarrow a + b = 20$ . $GM = 8 \Rightarrow ab = 64$ . Why this step? AM and GM package the sum and product — enough to rebuild the pair.

Step 3 — Solve the quadratic. $t^{2} - 20 t + 64 = 0 \Rightarrow t = \frac{20 \pm 400 - 256}{2} = \frac{20 \pm 12}{2} = 16 or 4$ . Why this step? $a, b$ are the roots of $t^{2} - (a + b) t + ab = 0$ .

Step 4 — Confirm HM of $16$ and $4$ . $\frac{2 \cdot 16 \cdot 4}{16 + 4} = \frac{128}{20} = 6.4$ . Matches Step 1. Why this step? Independent recomputation from the actual numbers seals the answer.

Verify: $a = 16, b = 4$ ; AM $= 10$ ✅, GM $= 64 = 8$ ✅, HM $= 6.4$ ✅, and $10 \geq 8 \geq 6.4$ obeys the chain. ✅ Forecast ("below 8") correct.

Recall Quick self-test on the matrix

Which cell forbids the reciprocal step entirely, and why? Answer: Cell D — a $0$ term has no reciprocal (that would need the undefined $\frac{1}{0}$ ), so no HP can contain $0$ .

In a genuinely sign-changing HP (Cell C), what sits at the crossover slot? Answer: a hole — the reciprocal-AP passes through $0$ there, and $\frac{1}{0}$ is undefined, so that term simply does not exist.

In Cell E part (b), why can $HM (4, b)$ never exceed $8$ ? Answer: as $b \to \infty$ , $H = \frac{8}{4/ b + 1} \to 8$ from below — HM is bounded by twice the smaller number.

In Cell F, why HM and not AM for the drone? Answer: equal distances make the times (reciprocals of speed) add, so the speeds combine harmonically.

In Cell H, which identity turns AM and GM straight into HM? Answer: $HM = \frac{GM ^{2}}{AM}$ .

Connections

Parent: Harmonic progression — definition, HM.
Arithmetic Progression — every example crosses here via the reciprocal bridge.
Geometric Progression, Geometric Mean, Arithmetic Mean — used together in Cell H.
Average speed and rates — the physical home of Cell F.
Resistors in parallel / Lens formula — same reciprocal-adding pattern in physics.