Harmonic Progression (HP): Formula, Mean, and Placement Exam Examples

Harmonic progression (HP) is a sequence whose reciprocals form an arithmetic progression. That one sentence contains the complete exam strategy: when you see an HP problem, take reciprocals, solve as AP, then invert.

What Harmonic Progression Is

If an AP has terms a, a+d, a+2d, ..., the corresponding HP is 1/a, 1/(a+d), 1/(a+2d), .... The numbers in the HP are not equally spaced; only their reciprocals are.

The nth-term formula follows directly:

• nth term of HP: T_n = 1 / (a + (n-1)d)

Here a is the first term and d is the common difference of the corresponding AP, not of the HP itself. You derive both by taking reciprocals of the given HP terms and treating the result as an AP.

The Brilliant.org harmonic progression reference covers the formal derivation and extended properties for those who want the proof.

Three numbers x, y, z are in HP when 1/x, 1/y, 1/z are in AP. The test: 2/y = 1/x + 1/z. This condition turns up in “are these three numbers in HP?” questions and in finding a middle term when the outer two are given.

In placement aptitude tests, HP appears in the sequences and series section alongside AP and GP. The structural differences are worth keeping clear: AP terms have a constant difference, GP terms have a constant ratio, and HP terms have a constant difference only after taking reciprocals. Examiners use this to set problems where students must first identify the progression type before choosing the right formula. Getting that identification step wrong wastes time on the correct arithmetic for the wrong sequence.

The Harmonic Mean

The harmonic mean (HM) of two numbers a and b is:

• HM(a, b): 2ab / (a + b)

For more than two numbers, the formula extends to:

• HM of n numbers: n / (1/a_1 + 1/a_2 + ... + 1/a_n)

Average Speed Application

When a vehicle covers the same distance at two different speeds, the true average speed is the harmonic mean of those speeds, not the arithmetic mean. Computing HM for speeds of 40 km/h and 60 km/h:

• HM = 2 × 40 × 60 / (40 + 60) = 4800 / 100 = 48 km/h

The arithmetic mean gives 50 km/h, which is wrong: the vehicle spends more time at the lower speed, so the true average is pulled below 50. HM accounts for this correctly.

AM, GM, and HM: Ordering

For any two positive numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM), which is always greater than or equal to the harmonic mean (HM). Written as an inequality: AM >= GM >= HM. The three means are equal only when both numbers are identical.

Placement tests occasionally ask for a calculation that requires knowing this ordering, or ask you to arrange three given means from largest to smallest. The direction is fixed: AM at the top, HM at the bottom.

Geometric Application

In any triangle, the inradius (radius of the inscribed circle) equals one-third of the harmonic mean of the three altitudes. If the altitudes are h_a, h_b, h_c:

• r = (1/3) × HM(h_a, h_b, h_c)

This identity appears in geometry MCQs that combine HP with basic mensuration.

Worked Placement Examples

The two question types below cover the bulk of what campus aptitude tests ask on HP. Both use the AP-conversion strategy.

Example 1: Find the 6th Term

Problem: The sum of the reciprocals of the first 11 terms of an HP is 110. Find the 6th term.

• Step 1: Reciprocals of HP terms form an AP: a, a+d, ..., a+10d.
• Step 2: Sum of 11 AP terms: S_11 = (11/2)(2a + 10d) = 11(a + 5d) = 110.
• Step 3: Solve: a + 5d = 10. This is the 6th AP term.
• Step 4: Invert: 6th HP term = 1/10.

Answer: 1/10

Example 2: Find the 16th Term

Problem: The 6th and 11th terms of an HP are 10 and 18. Find the 16th term.

• Step 1: Convert to AP. 6th HP term = 10 means 6th AP term = 1/10. 11th HP term = 18 means 11th AP term = 1/18.
• Step 2: Two equations:
  • a + 5d = 1/10
  • a + 10d = 1/18
• Step 3: Subtract the first from the second: 5d = 1/18 - 1/10 = -8/180 = -2/45. So d = -2/225.
• Step 4: Back-substitute: a = 1/10 - 5 × (-2/225) = 1/10 + 10/225 = 9/90 + 4/90 = 13/90.
• Step 5: 16th AP term = a + 15d = 13/90 + 15 × (-2/225) = 13/90 - 2/15 = 13/90 - 12/90 = 1/90.
• Step 6: Invert: 16th HP term = 90.

Answer: 90

Notice that d is negative in this example. A negative AP common difference means the AP terms are decreasing, which means the HP terms are increasing (a smaller positive denominator gives a larger fraction). Using this directional check before the final inversion is a fast way to catch sign errors.

The One-Step Solve Strategy

Every HP question in a campus placement test follows this structure:

1. Take reciprocals of every given HP term to obtain AP terms.
2. Identify a and d from those AP terms.
3. Find the required AP term: T_n = a + (n-1)d.
4. Invert: the HP term = 1 / T_n.

When the problem gives the sum of reciprocals (as in Example 1 above), that sum is the AP’s sum directly. Apply the AP sum formula S_n = (n/2)(2a + (n-1)d) and extract the required term.

Two errors that appear repeatedly in student working:

• Dividing by the wrong step gap: if two HP terms are k indices apart in the sequence, the reciprocal difference must be divided by k to get d. In Example 2, the 11th and 6th terms are 5 index steps apart, so the raw difference divides by 5.
• Inverting at the wrong step: invert only after finding the required AP term. Inverting the intermediate AP values produces the wrong HP terms and breaks every subsequent equation.

IndiaBix’s numbers and series practice set has additional problems on HP and related progressions. Drilling those alongside time and work problems builds the fraction arithmetic fluency that HP, mixture, and work-rate questions all draw on.

The campus placement evaluation test overview shows where HP fits in the broader quant syllabus. For a structured reading list, the best books for placement preparation page covers which texts handle sequences most clearly.

Systematic Thinking Past the Aptitude Round

The AP-conversion strategy works because it reduces HP to a known structure. That habit transfers. Most placement maths problems reward the same move: spot the underlying pattern, reduce to something already familiar.

Aptitude rounds are one gate. After clearing them, technical interviews and coding assessments follow. The step-by-step AP chain in Example 2 above (setting two equations, solving for d and a, then computing the 16th term) is the kind of systematic, multi-step reasoning that data-analytics shortlists like Mu Sigma’s aptitude test also reward. TinkerLLM at ₹299 is where engineers who have cleared those rounds go next: short, practical AI build projects rather than another video course.