Fundamentals of Business Mathematics and Statistics · Arithmetic Progression and Geometric Progression

Geometric Mean and the Relation between AM, GM and HM

Updated 10 October 2026 · Fact-checked

The geometric mean of two positive numbers a and b is √(ab). It is the middle term of a GP a, G, b. To insert n geometric means, find the common ratio r = (b ÷ a)^(1/(n+1)). For positive numbers, AM ≥ GM ≥ HM, and GM² = AM × HM.

Understand Geometric Mean and Relation between AM, GM and HM

A geometric progression (GP) is a sequence where each term is the previous term multiplied by a fixed number r, called the common ratio. Example: 3, 6, 12, 24 has r = 2.

If three numbers a, G, b are in GP, then G ÷ a = b ÷ G. This gives G² = ab, so G = √(ab). G is the geometric mean of a and b. It is defined here for positive a and b.

To insert n geometric means between a and b, you build a GP that starts with a and ends with b and has n terms in between. That GP has n + 2 terms. So b is the (n+2)th term, and b = a × r^(n+1). This gives r = (b ÷ a)^(1/(n+1)). The inserted means are ar, ar², ..., arⁿ.

For two positive numbers a and b, there are three famous means: AM = (a + b) ÷ 2, GM = √(ab), and HM = 2ab ÷ (a + b). They always satisfy AM ≥ GM ≥ HM. Equality holds only when a = b. Also, AM × HM = GM², so GM is the geometric mean of AM and HM.

AP versus GP: in an AP you add a constant difference; in a GP you multiply by a constant ratio. An AP grows by equal amounts, a GP grows by equal percentages.

Key formulas to remember

Geometric mean of two numbers: G = √(ab); For positive a and b. G is the middle term of the GP a, G, b.
nth term of a GP: Tₙ = a × r^(n−1); a is the first term and r is the common ratio.
Common ratio when inserting n means: r = (b ÷ a)^(1/(n+1)); The GP has n + 2 terms. The means are ar, ar², ..., arⁿ.
Product of n GMs inserted: G₁ × G₂ × ... × Gₙ = (ab)^(n/2); G₁, G₂, ..., Gₙ are the n inserted means. The product equals Gⁿ, where G = √(ab) is the single GM of a and b. Useful for quick checks.
Three terms in GP: b² = ac; If a, b, c are in GP (all non-zero), the middle term squared equals the product of the outer two.
Arithmetic mean and harmonic mean of two numbers: AM = (a + b) ÷ 2; HM = 2ab ÷ (a + b); HM is the reciprocal of the AM of the reciprocals.
Relation between AM, GM and HM: AM ≥ GM ≥ HM and GM² = AM × HM; For positive numbers. Equality holds only when a = b. The GM² relation is exact for two numbers.
Terms equidistant from the ends of a GP: T₁ × Tₙ = T₂ × Tₙ₋₁ = ...; Products of terms equally far from the two ends are equal.

How to solve Geometric Mean and Relation between AM, GM and HM questions

Use this method for any question on inserting means or on AM, GM and HM of two numbers.

1Identify what is asked: one GM, n GMs, or a relation among AM, GM and HM.
2For one GM between a and b, write G = √(ab). Check that a and b are positive.
3For n GMs, note the GP has n + 2 terms with first term a and last term b.
4Write b = a × r^(n+1) and solve r^(n+1) = b ÷ a. Take the root carefully.
5List the means as ar, ar², ..., arⁿ and verify that the last one times r gives b.
6For AM, GM, HM questions, compute any two and use GM² = AM × HM to find the third.
7Check the order AM ≥ GM ≥ HM on your numbers. If it fails, recheck the arithmetic.
8Match your answer to the option. Eliminate options that break the order or the GM² relation.

Quickest way: Shortcut with r and the GM² rule

When to use it: Use when the question gives two simple numbers and four numeric options.

Pick numbers that make a perfect power: for example 3 and 48 with 3 means gives r⁴ = 16, so r = 2.
Test options by multiplying: each mean must be the previous one times r, and the chain must end at b.
For AM and HM given, GM = √(AM × HM). Multiply first and then take the root.
The GM always lies between HM and AM, so reject any option for GM that is greater than the AM or less than the HM.

Common mistakes in Geometric Mean and Relation between AM, GM and HM

Using n instead of n + 1 in the root, so r = (b ÷ a)^(1/n).
Students forget the first and last numbers also count as terms.
Fix: Count terms: n means plus 2 ends gives n + 2 terms. The ratio is applied n + 1 times.
Taking the GM as (a + b) ÷ 2.
AM and GM formulas get mixed up.
Fix: GM involves a product and a root: √(ab). Say it as 'multiply, then root'.
Writing HM as (a + b) ÷ 2ab, which is upside down.
Memorising without linking it to reciprocals.
Fix: HM = 2ab ÷ (a + b). Check with a = b: HM must equal a.
Ignoring the sign when the root is even.
Students automatically take only the positive root.
Fix: For means between positive numbers, take the positive root. If a and b are both negative, the middle term of the GP a, G, b is −√(ab), because G must have the same sign as a and b. The AM ≥ GM ≥ HM relation applies only to positive numbers.
Applying AM ≥ GM ≥ HM to numbers of mixed signs or zero.
The rule is remembered without its condition.
Fix: The relation holds for positive numbers. Equality holds only when all numbers are equal.
Confusing AP and GP: adding the common ratio instead of multiplying.
Both topics use similar notation.
Fix: Ask: is the gap constant (AP) or the ratio constant (GP)? Test with the first three terms.

Worked examples

Example 1

Insert 3 geometric means between 3 and 48.

Show the solution

The GP has 3 + 2 = 5 terms. First term a = 3, fifth term = 48.
Use T₅ = a × r⁴. So 48 = 3 × r⁴, giving r⁴ = 16.
Therefore r = 2, taking the positive root.
The means are 3 × 2 = 6, 6 × 2 = 12, 12 × 2 = 24.
Check: 24 × 2 = 48, which matches the last term.

Answer: The three geometric means are 6, 12 and 24.

Example 2

The AM of two positive numbers is 25 and their HM is 16. Find their GM, and then find the two numbers.

Show the solution

Use GM² = AM × HM = 25 × 16 = 400.
So GM = √400 = 20.
Let the numbers be a and b. Then a + b = 2 × 25 = 50 and ab = GM² = 400.
The numbers are roots of x² − 50x + 400 = 0.
Factorise: (x − 10)(x − 40) = 0, so x = 10 or 40.
Check: AM = 25, GM = √400 = 20, HM = 2 × 400 ÷ 50 = 16. Order 25 ≥ 20 ≥ 16 holds.

Answer: GM = 20, and the numbers are 10 and 40.

Exam tips

Count terms first. Most errors in 'insert n means' come from using the wrong power in r.
Memorise GM² = AM × HM. It solves many two-number questions in one line.
Use the order AM ≥ GM ≥ HM to eliminate options. The GM must lie between the HM and the AM.
Use perfect powers (16 = 2⁴, 81 = 3⁴) to spot r fast. There is no negative marking, so always attempt every question.
For AP versus GP questions, check the difference or ratio of the first three terms before choosing.

Practice questions from Arithmetic Progression and Geometric Progression

Geometric Mean and Relation between AM, GM and HM: frequently asked questions

How do I insert n geometric means between two numbers?

Treat the numbers as the first and last terms of a GP with n + 2 terms. Find r = (b ÷ a)^(1/(n+1)). The means are ar, ar², ..., arⁿ.

Why is AM always at least GM, and GM at least HM?

For two positive numbers a and b, (a + b) ÷ 2 − √(ab) = (√a − √b)² ÷ 2, which is never negative. So AM ≥ GM. Since GM² = AM × HM and AM ≥ GM, we get HM = GM² ÷ AM ≤ GM. Equality holds only when a = b.

What is the difference between AP and GP?

In an AP, consecutive terms differ by a constant amount. In a GP, consecutive terms have a constant ratio. An AP uses addition and a GP uses multiplication.

Can the geometric mean be found for negative numbers?

The formula √(ab) gives the GM of two positive numbers. If a and b are both negative, the middle term of the GP a, G, b is −√(ab), because G must be negative like a and b. The relation AM ≥ GM ≥ HM applies only to positive numbers. For exam questions, assume positive numbers unless the question says otherwise.