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Arithmetic Sequences



Definition
Identifying Arithmetic Sequences
Finding Terms
Recursive Formula
Arithmetic Series
Arithmetic Mean
Arithmetic Sequences at a Glance



Constant Difference

An arithmetic sequence adds the same fixed value to each term to produce the next. That fixed value — the common difference — determines everything: whether the sequence rises or falls, how fast it grows, and what its partial sums look like. Arithmetic sequences are the simplest infinite pattern, and their summation formula is one of the oldest results in mathematics.



Definition

An arithmetic sequence is a sequence in which consecutive terms differ by a constant dd, called the common difference:

Common Difference
an+1−an=da_{n+1} - a_n = d
Learn more about this formula: Common Difference →

Starting from a first term a1a_1, the sequence unfolds as a1,a1+d,a1+2d,a1+3d,…a_1, a_1 + d, a_1 + 2d, a_1 + 3d, \ldots. The general term is:

General Term (Arithmetic Sequence)
an=a1+(n−1)da_n = a_1 + (n - 1)d
Learn more about this formula: General Term (Arithmetic Sequence) →


The common difference dd can be any real number. When d>0d > 0, the sequence increases: 3,7,11,15,…3, 7, 11, 15, \ldots has d=4d = 4. When d<0d < 0, it decreases: 20,17,14,11,…20, 17, 14, 11, \ldots has d=−3d = -3. When d=0d = 0, every term equals a1a_1 — a constant sequence, which is arithmetic in the trivial sense.

The explicit formula an=a1+(n−1)da_n = a_1 + (n-1)d has the same structure as a linear function f(x)=mx+bf(x) = mx + b: the common difference dd plays the role of slope, and the sequence's terms lie on a straight line when plotted against their indices.

Identifying Arithmetic Sequences

To determine whether a given sequence is arithmetic, compute the differences between consecutive terms. If every difference equals the same value, the sequence is arithmetic and that value is dd.

The sequence 5,12,19,26,335, 12, 19, 26, 33 has differences 7,7,7,77, 7, 7, 7. All equal, so it is arithmetic with d=7d = 7.

The sequence 1,4,9,16,251, 4, 9, 16, 25 has differences 3,5,7,93, 5, 7, 9. These are not constant, so the sequence is not arithmetic. (It is the sequence of perfect square numbers, which grows quadratically rather than linearly.)

The test applies to any number of terms: compute all consecutive differences and check for uniformity. A single pair of unequal differences is enough to disqualify. Conversely, if only a few terms are given, the test cannot confirm an arithmetic pattern with certainty — the sequence 2,4,62, 4, 6 could continue as 8,10,…8, 10, \ldots (arithmetic) or as 10,16,…10, 16, \ldots (something else). The rule generating the sequence, not just a finite sample, is what determines its type.

Finding Terms

The explicit formula an=a1+(n−1)da_n = a_1 + (n-1)d contains four quantities: ana_n, a1a_1, nn, and dd. Given any three, the fourth can be found by solving the equation.

Finding a specific term: if a1=4a_1 = 4 and d=6d = 6, then a20=4+19⋅6=118a_{20} = 4 + 19 \cdot 6 = 118.

Finding the common difference: if a1=3a_1 = 3 and a8=24a_8 = 24, then 24=3+7d24 = 3 + 7d, so d=3d = 3.

Finding the first term: if a12=50a_{12} = 50 and d=4d = 4, then 50=a1+11⋅450 = a_1 + 11 \cdot 4, so a1=6a_1 = 6.

Finding the position of a term: if a1=7a_1 = 7, d=5d = 5, and some term equals 5252, then 52=7+(n−1)⋅552 = 7 + (n-1) \cdot 5, so n−1=9n - 1 = 9 and n=10n = 10. The value 5252 is the 1010th term. If the equation yields a non-integer nn, the value does not appear in the sequence.

When two non-consecutive terms are given — say a3=11a_3 = 11 and a9=35a_9 = 35 — write both in terms of a1a_1 and dd: a1+2d=11a_1 + 2d = 11 and a1+8d=35a_1 + 8d = 35. Subtracting gives 6d=246d = 24, so d=4d = 4, and back-substituting gives a1=3a_1 = 3.
Given Find Equation / approach Example
a₁, d an an = a₁ + (n − 1)d — substitute directly a₁ = 4, d = 6 → a20 = 4 + 19·6 = 118
a₁, an, n d d = (an − a₁) ⁄ (n − 1) a₁ = 3, a₈ = 24 → d = 21 ⁄ 7 = 3
an, d, n a₁ a₁ = an − (n − 1)d a₁₂ = 50, d = 4 → a₁ = 50 − 44 = 6
a₁, d, a value v n (position of v) n = (v − a₁) ⁄ d + 1; if non-integer, v is not in the sequence a₁ = 7, d = 5, v = 52 → n = 45⁄5 + 1 = 10
Two non-consecutive terms a₁ and d solve the 2-equation linear system in a₁, d a₃ = 11, a₉ = 35 → 6d = 24 → d = 4, a₁ = 3

Recursive Formula

The recursive definition of an arithmetic sequence specifies a starting value and a rule for getting from one term to the next:

Recursive Form (Arithmetic Sequence)
a1=c,an=an−1+da_1 = c, \quad a_n = a_{n-1} + d
Learn more about this formula: Recursive Form (Arithmetic Sequence) →


Each term is built from the immediately preceding term by adding dd. This formulation makes the constant-difference structure explicit but requires computing all earlier terms to reach a later one.

Converting from recursive to explicit: if a1=ca_1 = c and an=an−1+da_n = a_{n-1} + d, then after applying the rule n−1n - 1 times, an=c+(n−1)da_n = c + (n-1)d. The explicit formula is a direct consequence of adding dd repeatedly.

Converting from explicit to recursive: given an=a1+(n−1)da_n = a_1 + (n-1)d, observe that an−an−1=da_n - a_{n-1} = d, which gives the recursive rule an=an−1+da_n = a_{n-1} + d with initial value a1a_1.

Both representations define the same sequence. The explicit form is better for computing isolated terms; the recursive form is better for understanding the building process and connects naturally to the idea of sequences defined by recurrence.

Arithmetic Series

The sum of the first nn terms of an arithmetic sequence is called an arithmetic series, denoted SnS_n:

Sn=a1+a2+a3+⋯+anS_n = a_1 + a_2 + a_3 + \cdots + a_n


The closed-form formula is:

Arithmetic Series Sum
Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n)
Learn more about this formula: Arithmetic Series Sum →


The derivation is straightforward. Write the sum forwards and backwards:

Sn=a1+(a1+d)+(a1+2d)+⋯+anS_n = a_1 + (a_1 + d) + (a_1 + 2d) + \cdots + a_n


Sn=an+(an−d)+(an−2d)+⋯+a1S_n = a_n + (a_n - d) + (a_n - 2d) + \cdots + a_1


Adding these two expressions term by term, every pair sums to a1+ana_1 + a_n, and there are nn such pairs:

2Sn=n(a1+an)2S_n = n(a_1 + a_n)


Dividing by 22 gives the result. An equivalent form replaces ana_n with a1+(n−1)da_1 + (n-1)d:

Sn=n2(2a1+(n−1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d)


This version is useful when ana_n is not known directly.

The sum of the first 100100 natural numbers, famously attributed to the young Gauss: a1=1a_1 = 1, a100=100a_{100} = 100, so S100=1002(1+100)=5050S_{100} = \frac{100}{2}(1 + 100) = 5050.

Arithmetic Mean

The arithmetic mean of two numbers aa and bb is:

Arithmetic Mean
M=a+b2M = \frac{a + b}{2}
Learn more about this formula: Arithmetic Mean →

In an arithmetic sequence, every term (except the first and last) is the arithmetic mean of its two neighbors:

Arithmetic Mean Property
an=an−1+an+12a_n = \frac{a_{n-1} + a_{n+1}}{2}
Learn more about this formula: Arithmetic Mean Property →


This follows directly from the constant-difference property: since an−an−1=an+1−an=da_n - a_{n-1} = a_{n+1} - a_n = d, adding the neighbors gives an−1+an+1=2ana_{n-1} + a_{n+1} = 2a_n.

Inserting arithmetic means between two values is the reverse problem: given endpoints aa and bb, place kk terms between them so that the resulting k+2k + 2 values form an arithmetic sequence. The common difference must be d=b−ak+1d = \frac{b - a}{k + 1}, and the inserted terms are a+d,a+2d,…,a+kda + d, a + 2d, \ldots, a + kd.

For example, inserting 33 arithmetic means between 55 and 2525 requires d=25−54=5d = \frac{25 - 5}{4} = 5, producing the sequence 5,10,15,20,255, 10, 15, 20, 25.

Arithmetic Sequences at a Glance

The page defined the common difference, derived the explicit and recursive formulas, the two equivalent series-sum formulas, the arithmetic-mean property, and the procedure for inserting means between two values. The table below collects each formula with a representative example for quick reference.
Concept Formula / property Example
Common difference d = an+1 − an (constant for every n) 5, 12, 19, 26 → d = 7
Explicit (n-th term) an = a₁ + (n − 1) · d a₁ = 4, d = 6 → an = 4 + 6(n − 1)
Recursive a₁ = c; an = an−1 + d a₁ = 4, an = an−1 + 6
Series sum (with an) Sn = n⁄2 · (a₁ + an) 1 + 2 + … + 100 = 100⁄2 · 101 = 5050
Series sum (with d) Sn = n⁄2 · (2a₁ + (n − 1) · d) used when an is not known directly
Arithmetic-mean property an = (an−1 + an+1) ⁄ 2 for interior terms every interior term is the mean of its two neighbors
Inserting k means between a, b d = (b − a) ⁄ (k + 1); inserts a + d, a + 2d, …, a + kd 3 means between 5, 25 → d = 5; 5, 10, 15, 20, 25
Linear-function structure an = d · n + (a₁ − d) — slope d, intercept (a₁ − d); plotted vs n the terms lie on a straight line d > 0 increasing, d < 0 decreasing, d = 0 constant