In short

A harmonic progression (HP) is a sequence of positive terms whose reciprocals form an arithmetic progression. If the reciprocals have first term a and common difference d, the n-th term of the HP is \dfrac{1}{a + (n-1)d}. Unlike APs and GPs, there is no neat closed-form formula for the sum of an HP — you handle HP problems by flipping to the reciprocal AP. Harmonic progressions appear naturally in physics (overtones of a vibrating string), in rate-and-work problems, and as the bridge to the harmonic mean.

A pipe fills a tank in 2 hours. A second pipe fills it in 3 hours. A third fills it in 6 hours. Their filling rates — the fraction of the tank filled per hour — are \dfrac{1}{2}, \dfrac{1}{3}, and \dfrac{1}{6}. Look at the denominators: 2, 3, 6. The differences are 1 and 3 — not constant. But now look at the reciprocals of the rates: 2, 3, 6. The reciprocals of those are \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{6} — which is where you started.

Try a different angle. Take the reciprocals of the original filling times: \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{6}. Now take their reciprocals: 2, 3, 6. Check: 3 - 2 = 1 and 6 - 3 = 3. Not an AP. So this particular sequence is not a harmonic progression.

But change the third pipe to one that fills the tank in 4 hours. Now the filling times are 2, 3, 4, which is an AP (common difference 1). The filling rates are \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4} — and this sequence is a harmonic progression, because its reciprocals (2, 3, 4) form an AP.

That is the definition of a harmonic progression: a sequence whose reciprocals are in AP.

A harmonic progression and its reciprocal APTwo rows of boxes. The top row shows the HP terms one-half, one-third, one-fourth, one-fifth. The bottom row shows the reciprocals 2, 3, 4, 5 which form an AP with common difference 1. Arrows labelled reciprocal connect the two rows. HP 1/2 1/3 1/4 1/5 ↕ ↕ ↕ ↕ reciprocal AP (d = 1) 2 3 4 5 +1 +1 +1
The HP $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}$ and its reciprocal AP $2, 3, 4, 5$. Every HP problem reduces to an AP problem: flip, solve, flip back.

Definition and the reciprocal trick

A sequence h_1, h_2, h_3, \dots of positive terms is a harmonic progression if the sequence of reciprocals

\frac{1}{h_1}, \; \frac{1}{h_2}, \; \frac{1}{h_3}, \; \dots

is an arithmetic progression. If the reciprocal AP has first term a and common difference d, then the n-th term of the reciprocal AP is a + (n-1)d, and the n-th term of the HP is

h_n = \frac{1}{a + (n - 1)d}

This formula is the entire machinery you need. There is no separate "HP formula" to memorise — you already know the AP formula, and the HP formula is its reciprocal.

Harmonic Progression

A sequence h_1, h_2, h_3, \dots of non-zero terms is a harmonic progression if

\frac{1}{h_1}, \; \frac{1}{h_2}, \; \frac{1}{h_3}, \; \dots

is an arithmetic progression. If the reciprocal AP has first term a and common difference d, the n-th term of the HP is

h_n = \frac{1}{a + (n-1)d}

The name "harmonic" comes from the physics of vibrating strings. A string vibrating at its fundamental frequency also vibrates at frequencies that are integer multiples of the fundamental — the overtones. The wavelengths of these overtones are L, \dfrac{L}{2}, \dfrac{L}{3}, \dfrac{L}{4}, \dots, which form an HP. Musicians call these the harmonics of the string, and the name stuck.

Vibrating string showing fundamental and first three overtonesFour horizontal panels showing a string fixed at both ends. The top panel shows the fundamental mode with one half-wave of wavelength L. The second panel shows the first overtone with two half-waves of wavelength L over 2. The third shows three half-waves at wavelength L over 3. The fourth shows four half-waves at wavelength L over 4. The wavelengths form an HP. λ = L λ = L/2 λ = L/3 λ = L/4 harmonics
A string of length $L$ vibrating in its fundamental mode and first three overtones. The wavelengths $L, \dfrac{L}{2}, \dfrac{L}{3}, \dfrac{L}{4}$ form an HP because their reciprocals $\dfrac{1}{L}, \dfrac{2}{L}, \dfrac{3}{L}, \dfrac{4}{L}$ (proportional to $1, 2, 3, 4$) form an AP.

Finding the n-th term

Since every HP problem is really an AP problem in disguise, the workflow is always the same three steps:

  1. Flip — take the reciprocals to get an AP.
  2. Solve — use the AP formula to find whatever you need.
  3. Flip back — take the reciprocal of the answer.

Take the HP \dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dfrac{1}{9}, \dots The reciprocals are 3, 5, 7, 9, \dots — an AP with first term a = 3 and common difference d = 2.

The n-th term of the reciprocal AP is 3 + (n-1) \times 2 = 2n + 1.

So the n-th term of the HP is \dfrac{1}{2n + 1}.

For n = 10: h_{10} = \dfrac{1}{2(10) + 1} = \dfrac{1}{21}.

HP terms shrinking toward zero as the reciprocal AP growsTwo bar charts stacked. The top chart shows the first seven HP terms one-third, one-fifth, one-seventh, and so on, as bars that shrink rapidly. The bottom chart shows the corresponding reciprocal AP terms 3, 5, 7, and so on, as bars that grow evenly. Arrows connect corresponding bars. HP terms (shrinking) 1/3 1/5 1/7 1/9 1/11 1/13 1/15 Reciprocal AP (growing) 3 5 7 9 11 13 15
The HP $\dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dots$ (red, shrinking) and its reciprocal AP $3, 5, 7, \dots$ (dark, growing). As the AP terms increase by a steady $+2$, the HP terms compress toward zero. The HP never reaches zero — it just gets arbitrarily close.

Why there is no direct sum formula

For an AP, the sum of the first n terms has a clean formula: S_n = \dfrac{n}{2}(2a + (n-1)d). For a GP, the sum is \dfrac{a(r^n - 1)}{r - 1}. But for an HP, no such closed-form formula exists.

The sum of the first n terms of the HP \dfrac{1}{1}, \dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dots is the harmonic series:

H_n = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{n}

There is no polynomial, exponential, or algebraic expression that gives H_n exactly. The best you can do is an approximation:

H_n \approx \ln n + \gamma

where \ln n is the natural logarithm and \gamma \approx 0.5772 is the Euler-Mascheroni constant. This approximation improves as n grows, but it is never exact.

The practical consequence: when an exam problem involves the sum of an HP, it is almost always asking you to find individual terms (using the reciprocal AP) or to use a specific property — not to compute the sum directly.

Interactive: partial sums of the harmonic series compared to ln(n)A coordinate plane with n on the horizontal axis from 1 to 30 and sum on the vertical axis from 0 to 5. Discrete dots show the partial sums H n of the harmonic series, which grow slowly. A smooth curve shows ln n plus 0.5772. The two are close but the dots are always slightly above the curve. A draggable point lets the reader explore values of n. 1 10 20 30 0 2 4 ln(n) + γ Hₙ (dots) drag to explore
The red dots show the partial sums $H_n = 1 + \tfrac{1}{2} + \cdots + \tfrac{1}{n}$. The smooth curve is $\ln n + \gamma$. The harmonic series grows, but painfully slowly — it takes about $e^{10} \approx 22026$ terms to reach a sum of $10$. Drag the point to see how close the approximation is.

Applications

Rate-and-work problems

The pipe-filling scenario from the opening is the classic HP application. If three workers can finish a job in a, b, and c hours respectively, their rates are \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c} jobs per hour. The combined rate is \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}, and the combined time is the reciprocal of that sum. When a, b, c are in AP, the rates are in HP — and the harmonic mean of the times gives the "average" time.

Resistors in parallel

In physics, when resistances R_1, R_2, \dots, R_n are connected in parallel, the total resistance satisfies

\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}

If the individual resistances form an AP, then the reciprocal resistances form an HP. The total resistance is related to the harmonic mean of the individual resistances.

Testing whether a sequence is in HP

Given a sequence, check whether it is in HP by taking reciprocals and testing whether the reciprocals form an AP. For example: is \dfrac{1}{5}, \dfrac{1}{8}, \dfrac{1}{11}, \dfrac{1}{14} an HP?

Reciprocals: 5, 8, 11, 14. Differences: 8 - 5 = 3, 11 - 8 = 3, 14 - 11 = 3. Constant difference, so the reciprocals form an AP with d = 3. The original sequence is an HP.

Testing HP by checking if reciprocals form an APA flowchart with three steps: Given sequence arrow to Take reciprocals arrow to Check constant difference. If the difference is constant, the output is HP confirmed. If not, the output is Not an HP. Given sequence Take reciprocals Check constant diff? HP confirmed
The HP test in three steps: take reciprocals, check for constant difference. If the reciprocals form an AP, the original sequence is an HP. This is the only test you need.

Two worked examples

Example 1: Find the 8th term of the HP $\dfrac{1}{3}, \dfrac{1}{7}, \dfrac{1}{11}, \dfrac{1}{15}, \dots$

Step 1. Take the reciprocals to form the AP.

3, 7, 11, 15, \dots

Why: every HP problem begins by flipping to the reciprocal AP.

Step 2. Identify the AP's parameters.

a = 3, \quad d = 7 - 3 = 4

Why: the first term of the AP is 3 and the common difference is the gap between consecutive terms.

Step 3. Find the 8th term of the AP.

a_8 = 3 + (8 - 1) \times 4 = 3 + 28 = 31

Why: the n-th term of an AP is a + (n-1)d.

Step 4. Flip back: the 8th term of the HP is

h_8 = \frac{1}{31}

Result: The 8th term of the HP is \dfrac{1}{31}.

Number line showing the reciprocal AP 3, 7, 11, 15 with the 8th term at 31A number line with tick marks at 3, 7, 11, 15, and 31. The first four points are filled circles. An ellipsis indicates intermediate terms. The 8th point at 31 is a red filled circle. Arrows between consecutive points are labelled plus 4. . . . 3 7 11 15 31 +4 +4 +4 n=1 n=2 n=3 n=4 n=8 Reciprocal AP: a₈ = 31, so HP term h₈ = 1/31
The reciprocal AP $3, 7, 11, 15, \dots$ with common difference $4$. The 8th term is $31$ (red), so the 8th term of the HP is $\dfrac{1}{31}$. The number line shows the AP growing in equal steps — the HP terms, being reciprocals of these, shrink toward zero.

The 8th HP term \dfrac{1}{31} is tiny compared to the first term \dfrac{1}{3}. As the reciprocal AP marches steadily to the right in equal steps of 4, the HP terms compress closer and closer to zero.

Example 2: If the 3rd and 7th terms of an HP are $\dfrac{1}{7}$ and $\dfrac{1}{15}$, find the HP.

Step 1. Flip to the reciprocal AP. The 3rd term of the AP is 7 and the 7th term is 15.

Why: the reciprocal of \dfrac{1}{7} is 7, and the reciprocal of \dfrac{1}{15} is 15.

Step 2. Use the two known AP terms to find a and d.

The n-th term of the AP is a + (n-1)d. From the two given terms:

a + 2d = 7 \quad \text{…(i)}
a + 6d = 15 \quad \text{…(ii)}

Why: the 3rd term uses n = 3, giving exponent n - 1 = 2; the 7th term uses n = 7, giving exponent 6.

Step 3. Subtract (i) from (ii):

4d = 8 \implies d = 2

Substitute back into (i): a + 4 = 7 \implies a = 3.

Why: subtracting eliminates a, leaving a single equation in d.

Step 4. Write the AP and flip back to the HP.

The reciprocal AP is 3, 5, 7, 9, 11, 13, 15, \dots

The HP is \dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dfrac{1}{9}, \dfrac{1}{11}, \dfrac{1}{13}, \dfrac{1}{15}, \dots

The general term is h_n = \dfrac{1}{3 + (n-1) \times 2} = \dfrac{1}{2n + 1}.

Result: The HP is \dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dfrac{1}{9}, \dots with general term h_n = \dfrac{1}{2n+1}.

The HP and its reciprocal AP shown side by side with known terms highlightedTwo rows. Top row: HP terms one-third, one-fifth, one-seventh highlighted in red, one-ninth, one-eleventh, one-thirteenth, one-fifteenth highlighted in red. Bottom row: AP terms 3, 5, 7 highlighted, 9, 11, 13, 15 highlighted. The 3rd and 7th terms are emphasised as the given data. HP 1/3 1/5 1/7 1/9 1/11 1/13 1/15 3rd 7th ↕ reciprocals AP (a = 3, d = 2) 3 5 7 9 11 13 15 Given the 3rd and 7th terms (red), the entire HP is determined
The HP $\dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \dots, \dfrac{1}{15}, \dots$ with the 3rd term ($\dfrac{1}{7}$) and 7th term ($\dfrac{1}{15}$) highlighted in red. Below it, the reciprocal AP $3, 5, 7, \dots, 15, \dots$ with the same terms highlighted. Two terms of the reciprocal AP are enough to determine the entire AP — and therefore the entire HP.

Two terms were enough to pin down the entire HP. The reciprocal AP has first term 3 and common difference 2. Every term of the HP follows from h_n = \dfrac{1}{2n+1}.

Common confusions

Going deeper

If you came here to learn the definition of an HP, how to find terms using the reciprocal AP, and where HPs appear, you have it. The rest of this section covers the harmonic series and a beautiful proof that it diverges.

The harmonic series diverges

The harmonic series H_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots + \dfrac{1}{n} grows without bound as n \to \infty. The growth is extremely slow — H_{100} \approx 5.19, H_{1000} \approx 7.49, H_{10000} \approx 9.79 — but it never levels off. No matter how large a target M you name, the harmonic series will eventually exceed M.

The proof groups the terms cleverly. After the first term (1) and the second term (\dfrac{1}{2}), group the next two terms:

\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}

Then group the next four terms:

\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}

Then the next eight terms also contribute more than \dfrac{1}{2}, and so on. Each doubling-group contributes at least \dfrac{1}{2} to the sum. Since there are infinitely many such groups, the sum exceeds every finite bound.

The relationship between HP, AP, and GP

If a, b, c are in AP, GP, and HP respectively, and all three means are of the same two positive numbers x and y, then:

a = \frac{x + y}{2}, \qquad b = \sqrt{xy}, \qquad c = \frac{2xy}{x + y}

and a beautiful relationship holds:

b^2 = ac

That is, the geometric mean of two numbers is also the geometric mean of their arithmetic mean and their harmonic mean. This identity links all three classical means and is explored fully in AM-GM-HM Inequality.

Where this leads next

The harmonic progression completes the trio of classical progressions. Combined with APs and GPs, it gives rise to the three classical means and their inequality chain.