In short

The arithmetic mean (AM) of two numbers a and b is \dfrac{a + b}{2} — the single number that sits exactly midway between them on the number line. More generally, inserting n arithmetic means between a and b creates an arithmetic progression of n + 2 terms, with common difference d = \dfrac{b - a}{n + 1}. The AM is always at least as large as the geometric mean of the same numbers — a fact that powers one of the deepest inequalities in mathematics.

A cricket team scores 246 in the first innings and 310 in the second. The commentator says the team's "average score" is 278. Where does that 278 come from? It is \dfrac{246 + 310}{2}, the number that is exactly as far above 246 as it is below 310. On a number line, 278 sits at the midpoint of the segment from 246 to 310 — a perfect balance point. If you replaced both scores with 278, the total would stay the same: 278 + 278 = 556 = 246 + 310.

That balance-point idea is the arithmetic mean. It is one of the oldest and most useful ideas in mathematics — and it connects directly to the arithmetic progressions you already know, because the AM of two numbers is nothing other than the middle term of a three-term AP.

The AM of two numbers

Take two numbers a and b. Their arithmetic mean is

A = \frac{a + b}{2}
The arithmetic mean as the midpoint of a segment on the number lineA horizontal number line with two marked points labelled a on the left and b on the right. A red point labelled A equals the quantity a plus b over two sits exactly halfway between them. Two equal-length arrows, both labelled d, show the distance from a to A and from A to b. a b A = (a + b)/2 d d
The arithmetic mean $A$ of two numbers $a$ and $b$ is the unique point equidistant from both. The two gaps, each of length $d = \dfrac{b - a}{2}$, are equal — so $a, A, b$ form a three-term AP with common difference $d$.

Three things to notice right away.

The AM is the midpoint. On the number line, A sits at the exact centre of the segment from a to b. The distance from a to A equals the distance from A to b, and both equal \dfrac{b - a}{2}.

The AM preserves the total. Replacing both numbers with their mean does not change the sum: A + A = a + b. This is the "fair share" interpretation — if two people have a and b rupees and they pool their money and split it equally, each gets A.

The AM makes a three-term AP. The sequence a,\; A,\; b is an arithmetic progression with common difference d = A - a = \dfrac{b - a}{2}. So "the AM of two numbers" and "the middle term of a three-term AP" are two names for the same thing.

Arithmetic Mean of two numbers

For any two real numbers a and b, the arithmetic mean is

A = \frac{a + b}{2}

The three numbers a, A, b form an arithmetic progression.

For the cricket example: a = 246, b = 310, and the AM is \dfrac{246 + 310}{2} = \dfrac{556}{2} = 278. The sequence 246, 278, 310 is an AP with common difference 32.

Inserting n arithmetic means

The idea extends naturally. Instead of one mean between a and b, you can insert n arithmetic means — creating an AP of n + 2 terms that starts at a and ends at b.

Take a = 3 and b = 27 with n = 3 means. You need a total of 3 + 2 = 5 terms: a,\; A_1,\; A_2,\; A_3,\; b. Since the five terms form an AP, the common difference is

d = \frac{b - a}{n + 1} = \frac{27 - 3}{3 + 1} = \frac{24}{4} = 6

So the three inserted means are:

A_1 = 3 + 6 = 9, \qquad A_2 = 3 + 12 = 15, \qquad A_3 = 3 + 18 = 21

and the full AP is 3,\; 9,\; 15,\; 21,\; 27.

Three arithmetic means inserted between 3 and 27A horizontal number line with five evenly spaced points at 3, 9, 15, 21, and 27. The endpoints 3 and 27 are drawn as filled circles. The three intermediate points 9, 15, and 21 are drawn as red filled circles indicating they are the inserted arithmetic means. Arrows between consecutive points are each labelled plus 6, showing the common difference. 3 9 15 21 27 +6 +6 +6 +6
Three arithmetic means (red) inserted between $3$ and $27$. The five terms $3, 9, 15, 21, 27$ form an AP with common difference $d = 6$. The means divide the segment from $3$ to $27$ into four equal steps.

The general formula: to insert n arithmetic means between a and b, the common difference of the resulting AP is

d = \frac{b - a}{n + 1}

and the k-th inserted mean (for k = 1, 2, \dots, n) is

A_k = a + k \cdot d = a + k \cdot \frac{b - a}{n + 1}

The logic is clean. You have n + 2 terms in total (the two endpoints plus the n means), so there are n + 1 gaps between consecutive terms. Since the total span is b - a, each gap must be \dfrac{b - a}{n + 1}.

The sum of n arithmetic means

Here is a useful property. The sum of all n inserted means equals n times the simple AM of a and b:

A_1 + A_2 + \cdots + A_n = n \cdot \frac{a + b}{2}

The proof uses the AP sum formula. The n + 2 terms form an AP with first term a and last term b, so their sum is \dfrac{(n + 2)(a + b)}{2}. Subtract the two endpoints to get the sum of the n means:

\frac{(n + 2)(a + b)}{2} - a - b = \frac{(n + 2)(a + b)}{2} - \frac{2(a + b)}{2} = \frac{n(a + b)}{2}

which is indeed n \cdot \dfrac{a + b}{2}.

Properties of the arithmetic mean

The AM has several properties that make it a natural "centre" for a set of numbers.

Property 1: The AM lies between the extremes. For any two numbers a \leq b, the AM satisfies a \leq \dfrac{a + b}{2} \leq b. The mean never falls outside the range of its inputs.

Property 2: Equal deviations. The sum of the deviations from the mean is zero. For two numbers:

(a - A) + (b - A) = a + b - 2A = a + b - (a + b) = 0

This extends to any number of values. If x_1, x_2, \dots, x_n have mean \bar{x} = \dfrac{x_1 + x_2 + \cdots + x_n}{n}, then

\sum_{i=1}^{n} (x_i - \bar{x}) = 0

The positive deviations (values above the mean) exactly cancel the negative deviations (values below). This is what "balance point" means quantitatively.

Deviations from the mean cancel outA number line showing five data points at positions 2, 5, 6, 9, and 13. A red vertical line marks their arithmetic mean at 7. Arrows above the line show each point's deviation from the mean: negative 5, negative 2, negative 1, positive 2, and positive 6. The sum of these deviations is zero. 2 5 6 9 13 mean = 7 −5 −2 −1 +2 +6 (−5) + (−2) + (−1) + (+2) + (+6) = 0
Five values — $2, 5, 6, 9, 13$ — have mean $\dfrac{35}{5} = 7$. The deviations from the mean are $-5, -2, -1, +2, +6$, and they sum to zero. The mean is the balance point of the data: the leftward pull equals the rightward pull.

Property 3: The AM of an AP is the middle term. In any AP with an odd number of terms, the arithmetic mean of all the terms equals the middle term. In an AP with an even number of terms, it equals the average of the two middle terms. This follows from the symmetry of an AP around its centre.

Property 4: Changing one value. If one of n values increases by c, the mean increases by \dfrac{c}{n}. The change in the mean is always a scaled-down version of the change in the data. This is because the mean distributes any change evenly across all values.

The AM for more than two numbers

The cricket commentator uses the AM of two innings. But a batsman's "batting average" over an entire season uses the same idea with more numbers. If the scores are x_1, x_2, \dots, x_n, the arithmetic mean is

\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} = \frac{1}{n}\sum_{i=1}^{n} x_i

The formula says: add everything up, then divide by the count. All the properties above — balance point, equal deviations, preserving the total — carry over unchanged.

Interactive: drag the point and watch the mean shiftA number line from 0 to 20 with four fixed data points at 4, 7, 10, and 15. A fifth red draggable point starts at 9. A red vertical bar marks the arithmetic mean of all five values. As the reader drags the fifth point, the mean shifts in real time. 0 5 10 15 20 4 7 10 15 ↔ drag the red point
Four fixed scores ($4, 7, 10, 15$) plus one draggable value. The readout shows the current mean of all five. Drag the red point and watch the mean respond — when you move the point by $5$ units, the mean shifts by $\dfrac{5}{5} = 1$ unit, exactly as Property 4 predicts.

Two worked examples

Example 1: Find 4 arithmetic means between 8 and 33

The problem asks for a sequence 8,\; A_1,\; A_2,\; A_3,\; A_4,\; 33 — six terms forming an AP.

Step 1. Count the gaps. There are 4 + 2 = 6 terms, so 6 - 1 = 5 gaps.

Why: n arithmetic means between two endpoints create n + 2 terms total, and an AP of n + 2 terms has n + 1 gaps.

Step 2. Find the common difference.

d = \frac{33 - 8}{5} = \frac{25}{5} = 5

Why: the total span b - a = 25 must be divided equally among the 5 gaps.

Step 3. List the means.

A_1 = 8 + 5 = 13
A_2 = 8 + 10 = 18
A_3 = 8 + 15 = 23
A_4 = 8 + 20 = 28

Why: each mean is a + kd for k = 1, 2, 3, 4.

Step 4. Verify: the full sequence is 8, 13, 18, 23, 28, 33. Each consecutive difference is 5, and the last term is 33. Also check the sum property: 13 + 18 + 23 + 28 = 82, and 4 \times \dfrac{8 + 33}{2} = 4 \times 20.5 = 82. Confirmed.

Result: The four arithmetic means are 13, 18, 23, 28.

Six-term AP from 8 to 33 with common difference 5A number line showing six evenly spaced points at 8, 13, 18, 23, 28, and 33. The endpoints 8 and 33 are black circles. The four intermediate points are red circles representing the four inserted arithmetic means. Each gap is labelled plus 5. 8 13 18 23 28 33 +5 +5 +5 +5 +5
The six-term AP $8, 13, 18, 23, 28, 33$ with common difference $d = 5$. The four red points are the arithmetic means, dividing the segment from $8$ to $33$ into five equal steps. The figure is the picture version of the formula $A_k = 8 + 5k$.

Example 2: The mean of five observations is 12. If one observation valued 20 is removed, find the new mean.

This is a "change in the mean" problem. Instead of inserting means into an AP, you are working with a data set and using the total-preserving property.

Step 1. Find the total of the five observations.

\text{total} = 5 \times 12 = 60

Why: the mean is the total divided by the count. Multiplying the mean by the count recovers the total.

Step 2. Remove the observation valued 20.

\text{new total} = 60 - 20 = 40

Why: removing one value subtracts it from the total.

Step 3. Find the new count and the new mean.

\text{new count} = 5 - 1 = 4
\text{new mean} = \frac{40}{4} = 10

Why: four values remain, and their total is 40.

Step 4. Check the logic. The removed observation (20) was above the original mean (12), so removing it should pull the mean down — and indeed the new mean (10) is lower than the old mean (12). The direction makes sense.

Result: The new mean is 10.

Removing a value above the mean pulls the mean downTwo horizontal bars. The top bar has width proportional to 60, representing the total of five observations with mean 12. A segment of width 20 at the right end is marked as the removed observation. The bottom bar has width proportional to 40, representing the total of four remaining observations with mean 10. A vertical red line marks the mean on each bar. remaining: 40 removed: 20 total: 60 5 values, mean = 12 remaining: 40 total: 40 4 values, mean = 10 old mean: 12 new mean: 10
The top bar represents the total ($60$) of five observations with mean $12$. The shaded segment ($20$) is the removed observation. After removal, the bottom bar has total $40$ across four values, giving a new mean of $10$. Since the removed value was above the mean, the mean drops — visible as the red line shifting to the left.

Common confusions

Going deeper

If you came here to learn what the arithmetic mean is, how to insert means, and how the properties work, you have it. The rest of this section is for readers who want to see how the AM connects to deeper ideas — the AM-GM inequality, weighted means, and why the AM minimises a specific measure of spread.

The AM-GM inequality

Take two positive numbers a and b. Their arithmetic mean is \dfrac{a + b}{2}. Their geometric mean is \sqrt{ab}. The AM-GM inequality says:

\frac{a + b}{2} \geq \sqrt{ab}

with equality if and only if a = b.

The proof for two numbers is short. Start from the fact that any real square is non-negative:

(\sqrt{a} - \sqrt{b})^2 \geq 0

Expand:

a - 2\sqrt{ab} + b \geq 0

Rearrange:

\frac{a + b}{2} \geq \sqrt{ab}

Equality holds precisely when \sqrt{a} = \sqrt{b}, that is, when a = b. The AM always sits at or above the GM, and it touches the GM only when both numbers are equal. The article on AM-GM-HM Inequality develops this into a chain of inequalities linking three different means.

Weighted arithmetic mean

In a class of 30 students, 20 score an average of 70 marks and 10 score an average of 90 marks. The overall class average is not \dfrac{70 + 90}{2} = 80 — it is the weighted average:

\bar{x} = \frac{20 \times 70 + 10 \times 90}{20 + 10} = \frac{1400 + 900}{30} = \frac{2300}{30} \approx 76.7

The group with more students pulls the mean closer to its score. The formula is

\bar{x} = \frac{w_1 x_1 + w_2 x_2 + \cdots + w_k x_k}{w_1 + w_2 + \cdots + w_k}

where w_i are the weights (here, the group sizes). The ordinary AM is the special case where all weights are equal.

The AM minimises sum of squared deviations

Here is a fact that explains why the arithmetic mean shows up everywhere in statistics. Among all possible choices of a single number c, the value that minimises

\sum_{i=1}^{n} (x_i - c)^2

is exactly c = \bar{x}, the arithmetic mean. The AM is the number that is, on average, closest to every value in the data set — "closest" in the sense of minimising the sum of squared distances. This is the mathematical reason the AM is the default "centre" in statistics, and it connects directly to the idea of variance and standard deviation.

Where this leads next

The arithmetic mean is one of three classical means. The other two — geometric and harmonic — measure centrality in different ways, and the relationship among all three is one of the cornerstones of inequality theory.