Harmonic Progressions
On this page
Direct answer
Reciprocals of a harmonic progression sit in arithmetic progression: 1/a, 1/(a + d), 1/(a + 2d), ... — so every HP technique is an AP technique applied after inversion, and the first move in any HP problem is to flip the terms. There is no general closed-form sum of harmonic terms, hence questions drill terms and means. The harmonic mean of two numbers is HM = 2ab/(a + b); for three terms a, b, c in HP the middle satisfies b = 2ac/(a + c). Two anchors: GM² = AM × HM for two positive numbers, and the impossibility that three distinct terms in AP can also be in HP — the structures coexist only in a constant sequence.
What you must remember
- Definition by inversion: a, b, c are in HP exactly when 1/a, 1/b, 1/c are in AP, i.e. 2/b = 1/a + 1/c, i.e. b = 2ac/(a + c).
- nth term: 1/(a + (n − 1)d) where a and d belong to the reciprocal AP — compute the reciprocal's nth term, then flip.
- Harmonic mean: HM of two numbers is 2ab/(a + b); of n numbers, n divided by the sum of reciprocals.
- Three-number setup: take the reciprocals as a − d, a, a + d to exploit symmetry — but remember the d = 0 collapse below.
- The mean chain: for two positive numbers, AM ≥ GM ≥ HM with GM² = AM × HM — the equality holding only when the numbers coincide.
- No sum formula: Σ 1/k has no elementary closed form; any "sum of an HP" question is either a misreading or a telescoping product in disguise.
- AP-HP collision: a − d, a, a + d in HP forces d = 0 — verify by the 2/b condition — so no non-constant sequence is both.
Terms without sums
An HP has second term 1/4 and sixth term 1/12; find its tenth term. Flip first: the reciprocal AP has second term 4 and sixth term 12, so (a + 5d) − (a + d) = 4d = 8, giving d = 2 and a = 2. The tenth reciprocal is a + 9d = 20, so the tenth HP term is 1/20. Notice the discipline: nothing was ever added across harmonic terms — the AP did every computation, and each answer returned through one reciprocal. Now the collision result as a worked warning: try to find three terms in both AP and HP by writing a − d, a, a + d and imposing 2/a = 1/(a − d) + 1/(a + d) = 2a/(a² − d²); this forces a² − d² = a², hence d = 0. The only sequences living in both progressions are constants — a one-line proof that makes a satisfying multiple-correct statement, and a reminder that symmetry setups with nonzero d can never satisfy the harmonic condition.
Slips that cost terms
JEE Main asks nth terms, harmonic means between numbers, and the AM-GM-HM inequalities — where the standard leak is computing the HM of a and b as (a + b)/2 (the AM) or as √(ab) (the GM) under time pressure; the HM is the reciprocal of the average of reciprocals, and the phrase "harmonic" must trigger the flip before anything else. A recurring Main pattern hides the HP inside another structure: two numbers a and b have harmonic mean 12/5? Then 2ab/(a + b) = 12/5 combined with, say, the GM produces the pair. JEE Advanced enjoys the collision territory: statements like "if a, b, c are in HP then 1/(b + c), 1/(c + a), 1/(a + b) are in AP" — testable by pure algebra after inversion — and the GM² = AM × HM identity used to build equations. The sum temptation is the designated trap: any option offering a closed form for a harmonic sum is wrong by construction, and recognising it instantly beats any computation. Flip first, compute in the AP, flip back, and never add harmonic terms directly.
Frequently asked questions
What defines a harmonic progression?
A sequence whose reciprocals form an arithmetic progression; a, b, c are in HP exactly when b = 2ac/(a + c).
How do you find the nth term of an HP?
Write the reciprocal AP's nth term a + (n − 1)d with the common difference of reciprocals, then take its reciprocal.
What is the harmonic mean of two numbers?
2ab/(a + b) — the reciprocal of the arithmetic mean of the reciprocals; for n numbers it is n divided by the sum of reciprocals.
Why is there no sum formula for harmonic progressions?
The partial sums of 1/k have no elementary closed form, unlike AP and GP — so examinable HP questions concern terms and means, never sums.
Can three distinct numbers be in both AP and HP?
No: writing a − d, a, a + d and imposing the harmonic condition forces d = 0, so only constant sequences belong to both.