In short

An arithmetic progression (AP) is a sequence in which every consecutive pair of terms has the same difference d, called the common difference. If the first term is a and the common difference is d, the n-th term is a_n = a + (n-1)d. This formula turns the recursive pattern "add d each time" into a direct formula that gives you any term instantly. APs are linear — their terms lie on a straight line when plotted — and they show up whenever something grows (or shrinks) by a fixed amount at each step.

An auto-rickshaw meter in a city starts at ₹25 for the first kilometre, then adds ₹12 for each additional kilometre. After 1 km the fare is ₹25. After 2 km, ₹37. After 3 km, ₹49. After 4 km, ₹61. Write these fares in order:

25, \; 37, \; 49, \; 61, \; 73, \; \dots

Check the gaps: 37 - 25 = 12, 49 - 37 = 12, 61 - 49 = 12. Every consecutive pair differs by the same amount — ₹12. That constant gap is the defining feature of an arithmetic progression. The fare after n kilometres is not some mysterious formula; it is just the starting fare plus (n-1) copies of ₹12. The tenth kilometre? 25 + 9 \times 12 = 133. No surprises.

This article is about that structure: sequences with a constant difference. You will see exactly why the n-th term formula works, how to recognise an AP, and what properties make it a workhorse in mathematics.

Rickshaw fare as an arithmetic progressionFive boxes in a row showing the fares 25, 37, 49, 61, 73 rupees after 1 through 5 kilometres. Between each pair of boxes an arrow is labelled plus 12, showing the constant difference. ₹25 1 km ₹37 2 km ₹49 3 km ₹61 4 km ₹73 5 km +12 +12 +12 +12 Each fare is exactly ₹12 more than the one before — a constant gap.
The rickshaw fare sequence $25, 37, 49, 61, 73, \dots$ is an arithmetic progression with first term $a = 25$ and common difference $d = 12$. The constant gap is what makes it an AP.

Definition and common difference

Arithmetic Progression

A sequence a_1, a_2, a_3, \dots is an arithmetic progression (AP) if there exists a constant d such that

a_{n+1} - a_n = d \quad \text{for all } n \geq 1.

The constant d is called the common difference. The first term is usually denoted a.

The common difference can be positive (the sequence increases), negative (the sequence decreases), or zero (every term is the same).

AP First term a Common difference d Behaviour
3, 7, 11, 15, \dots 3 4 increasing
20, 17, 14, 11, \dots 20 -3 decreasing
5, 5, 5, 5, \dots 5 0 constant

A constant sequence is a special case of an AP — the most boring one, but still legitimate.

To check whether a given sequence is an AP, compute consecutive differences a_2 - a_1, a_3 - a_2, a_4 - a_3, and so on. If they are all equal, it is an AP and that common value is d. If even one difference disagrees, it is not an AP.

Take 2, 6, 18, 54, \dots The differences are 4, 12, 36 — not constant. This is not an AP. (It is a geometric progression — the ratio between consecutive terms is constant, but that is a different story.)

Three APs with positive, negative, and zero common differenceThree rows of plotted points on a coordinate grid. The top row rises from left to right with d equals 4. The middle row falls from left to right with d equals negative 3. The bottom row is flat with d equals 0. Each row has five points connected by a dashed line. n aₙ 1 2 3 4 5 3 7 11 15 19 d = 4 d = −3 d = 0
Three APs plotted on the same grid. The red dots ($d = 4$) climb; the grey dots ($d = -3$) descend; the muted dots ($d = 0$) stay flat. All three sets of points sit on straight lines — the hallmark of an arithmetic progression.

The n-th term formula — derivation

You know the first term a and the common difference d. What is the n-th term?

Build it step by step. Each term is the previous one plus d:

a_1 = a
a_2 = a + d
a_3 = a + d + d = a + 2d
a_4 = a + d + d + d = a + 3d

The pattern: the n-th term has (n - 1) copies of d added to the first term a. Why n - 1 and not n? Because the first term is the starting point — you add d zero times to get a_1, once to get a_2, twice to get a_3, and so on. By the time you reach a_n, you have added d exactly n - 1 times.

a_n = a + (n - 1)d

That is the formula. It is worth pausing to see why it must be right. The sequence is defined by a_1 = a and a_{n+1} = a_n + d. By mathematical induction, if a_n = a + (n-1)d, then a_{n+1} = a_n + d = a + (n-1)d + d = a + nd = a + ((n+1) - 1)d. The formula holds for a_1 = a + 0 \cdot d = a (base case), and the inductive step carries it forward. So it holds for all n.

Building the nth term of an AP one step at a timeA horizontal number line starting at the point a. Five arrows of equal length d point rightward, each one landing on the next term. The first arrow goes from a to a plus d, the second from a plus d to a plus 2d, and so on up to a plus 4d. Below the line, labels show that after n minus 1 jumps you reach a plus n minus 1 times d. a a₁ +d a+d a₂ +d a+2d a₃ +d ··· a+(n−1)d aₙ (n − 1) jumps of size d aₙ = a + (n − 1)d
Starting from $a$, each jump of size $d$ lands on the next term. After $n - 1$ jumps, you are at $a + (n-1)d$ — the $n$-th term of the AP.

The formula also reveals why AP terms lie on a straight line when plotted. Expand it:

a_n = a + (n-1)d = dn + (a - d)

This is a linear function of n, of the form a_n = dn + c where c = a - d. The slope is d (the common difference) and the y-intercept is a - d. A sequence is an AP if and only if its terms, when plotted against their index, fall on a straight line.

Drag the red point along the line below to pick any term of the AP with a = 3 and d = 4. The readout shows the index n and the value a_n = 4n - 1 — the formula in action.

Interactive: drag a point along an arithmetic progressionA coordinate plane with the line y equals 4x minus 1 plotted from x equals 1 to 12. A draggable red point sits on the line. A readout shows the current index n and the computed term value a sub n equals 4n minus 1. n aₙ 2 3 4 5 6 7 8 9 10 ↔ drag to pick a term
Drag the point to see the $n$-th term formula at work. At $n = 1$ the term is $3$; at $n = 10$ it is $39$. The straight line is the visual signature of the constant common difference $d = 4$.

Properties of an AP

Several useful properties follow directly from the n-th term formula.

Property 1: Any three consecutive terms satisfy a_{n+1} = \frac{a_n + a_{n+2}}{2}.

In other words, the middle term of any three consecutive terms is the average of the outer two. Check it: a_n = a + (n-1)d, a_{n+2} = a + (n+1)d. Their average is \frac{(a + (n-1)d) + (a + (n+1)d)}{2} = \frac{2a + 2nd}{2} = a + nd = a_{n+1}.

This property also works backwards: if three numbers p, q, r satisfy q = \frac{p + r}{2}, then p, q, r are in AP. The number q is called the arithmetic mean of p and r.

Property 2: The n-th term from the end.

If an AP has m terms total, the n-th term from the end is a_m - (n-1)d, where a_m is the last term. This is the same formula run backwards, with -d replacing d.

Property 3: Sum of terms equidistant from the ends.

In a finite AP with first term a and last term l, the sum of the first and last terms equals the sum of the second and second-to-last, equals the sum of the third and third-to-last, and so on:

a_1 + a_m = a_2 + a_{m-1} = a_3 + a_{m-2} = \dots = a + l

Why? a_k + a_{m+1-k} = [a + (k-1)d] + [a + (m-k)d] = 2a + (m-1)d = a + l for every k. Each pair sums to the same value.

This property is the key to deriving the sum formula in the next article.

Equidistant terms in an AP have equal sumsA row of seven boxes representing an AP with terms a1 through a7. Curved arches connect a1 with a7, a2 with a6, and a3 with a5. Each arch is labelled a plus l, showing that the sum of terms equidistant from the ends is the same constant. a₁ a₂ a₃ a₄ a₅ a₆ a₇ a + l a + l a + l Every pair of terms equidistant from the ends sums to a + l.
In a seven-term AP, the pairs ($a_1, a_7$), ($a_2, a_6$), and ($a_3, a_5$) each sum to $a + l$. The middle term $a_4$ is both the average of the sequence and the midpoint between first and last.

Property 4: Inserting arithmetic means.

If you have two numbers p and q and want to place k numbers between them so that the whole list is in AP, those inserted numbers are called k arithmetic means between p and q. The common difference of the resulting AP is d = \frac{q - p}{k + 1}, and the r-th inserted number is p + r \cdot \frac{q - p}{k + 1}.

For instance, to insert 3 arithmetic means between 2 and 18: d = \frac{18 - 2}{3 + 1} = 4, so the three means are 6, 10, 14, and the full AP is 2, 6, 10, 14, 18.

Worked examples

Example 1: Find the 20th term and the general term of the AP $7, 3, -1, -5, \dots$

Step 1. Identify a and d.

The first term is a = 7. The common difference is d = 3 - 7 = -4.

Why: the common difference is always a_2 - a_1. Here it is negative, so the sequence decreases.

Step 2. Write the general term.

a_n = a + (n - 1)d = 7 + (n - 1)(-4) = 7 - 4n + 4 = 11 - 4n

Why: substitute a = 7 and d = -4 into the formula and simplify.

Step 3. Plug in n = 20.

a_{20} = 11 - 4(20) = 11 - 80 = -69

Why: the general term gives random access — you do not need to list 19 previous terms.

Step 4. Verify with a known term.

a_3 = 11 - 4(3) = 11 - 12 = -1. The third term in the given sequence is -1. Correct.

Why: checking against a known term catches sign errors and off-by-one mistakes.

Result: The general term is a_n = 11 - 4n, and a_{20} = -69.

The decreasing AP 7, 3, negative 1, negative 5, negative 9 plotted for 5 termsA coordinate plane with n from 1 to 7 on the horizontal axis and a sub n from negative 9 to 7 on the vertical axis. Five red dots are plotted along a descending straight line. The dots cross below zero between n equals 2 and n equals 3. n aₙ 1 2 3 4 5 6 7 7 3 0 −1 −5 −9
The AP $7, 3, -1, -5, -9, \dots$ plotted on a coordinate plane. The points descend in a straight line, crossing zero between $n = 2$ and $n = 3$. The negative slope reflects the negative common difference $d = -4$.

The graph shows the line going down and to the right — exactly what a negative common difference looks like. By n = 20, the line has descended far below zero to a_{20} = -69.

Example 2: The 5th and 11th terms of an AP are $19$ and $43$. Find $a$, $d$, and $a_{15}$.

Step 1. Set up two equations from the given information.

a_5 = a + 4d = 19 and a_{11} = a + 10d = 43.

Why: the n-th term formula a_n = a + (n-1)d applied at n = 5 and n = 11 gives two linear equations in two unknowns.

Step 2. Subtract the first equation from the second.

(a + 10d) - (a + 4d) = 43 - 19, giving 6d = 24, so d = 4.

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

Step 3. Substitute back to find a.

a + 4(4) = 19, so a = 19 - 16 = 3.

Why: with d known, either equation gives a directly.

Step 4. Compute a_{15}.

a_{15} = 3 + 14 \times 4 = 3 + 56 = 59.

Why: plug n = 15, a = 3, d = 4 into the formula.

Result: a = 3, d = 4, a_{15} = 59.

Finding an AP from two known termsA number line from n equals 1 to n equals 15. Three points are highlighted: (5, 19), (11, 43), and (15, 59). A straight line connects all points. Labels show the given terms at positions 5 and 11 and the computed term at position 15. n aₙ 0 5 11 15 19 given 43 given 59 found
Two given terms ($a_5 = 19$ and $a_{11} = 43$) fix a unique straight line. The line passes through every term of the AP — reading off $n = 15$ gives $a_{15} = 59$. Two points determine a line, and two terms of an AP determine the entire AP.

Two terms of an AP determine everything, because two points determine a line. Once you know any two terms and their positions, the first term, the common difference, and every other term follow.

Common confusions

Going deeper

If you came here to understand what an AP is, derive the n-th term formula, and see the key properties, you have everything you need. The rest of this section is for readers who want the connection to linear functions and a few less obvious properties.

APs and linear functions

The n-th term of an AP is a_n = dn + (a - d). Compare this with the slope-intercept form of a linear function: f(x) = mx + c. The common difference d plays the role of slope m, and a - d plays the role of the y-intercept c.

This is not an analogy — it is an exact correspondence. An infinite sequence is an AP if and only if the function f(n) = a_n is a linear function of n. That is why the plotted terms of an AP always fall on a straight line.

The correspondence goes further. The problem "find d given two terms" is exactly the problem "find the slope of a line given two points." The problem "find a_n for a specific n" is exactly "evaluate a linear function at a point." Anything you know about linear equations transfers directly to APs.

If a_n is a linear polynomial in n, the sequence is an AP

Conversely, any sequence whose general term is a polynomial of degree 1 in n — that is, a_n = pn + q for constants p and q — is an AP with first term a_1 = p + q and common difference d = p. Check: a_{n+1} - a_n = [p(n+1) + q] - [pn + q] = p, which is constant.

This is the only polynomial degree that gives an AP. A quadratic general term like a_n = n^2 gives differences 3, 5, 7, 9, \dots that are not constant — the sequence of perfect squares is not an AP. (But the differences form an AP — a fact that leads to the theory of finite differences.)

Three numbers in AP

A useful shorthand for competition and exam problems: if three numbers are in AP, write them as a - d, \; a, \; a + d. Their sum is 3a (the d terms cancel), which often lets you find a immediately. Their product or any other symmetric expression simplifies because of the cancellation.

For four numbers in AP, write them as a - 3d, \; a - d, \; a + d, \; a + 3d (spacing by 2d). The sum is 4a.

This trick — centring the AP at its middle term — appears in almost every problem where you need to find three or four terms in AP with given conditions.

Where this leads next