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Arithmetic Sequences



Definition
Sequence Notation
Identifying Arithmetic Sequences
Finding Terms
Recursive Formula
Arithmetic Series
Arithmetic Mean
Arithmetic Sequences at a Glance
Common Pitfalls
Arithmetic Sequences FAQ



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

Six number patterns. In which of them does every step add the same amount?
A: 3,7,11,153, 7, 11, 15 \quad B: 20,17,14,1120, 17, 14, 11 \quad C: 1,4,9,161, 4, 9, 16 \quad D: 2,4,8,162, 4, 8, 16 \quad E: 5,5,5,55, 5, 5, 5 \quad F: 1,1.5,2,2.51, 1.5, 2, 2.5
Subtracting each term from the next settles it: A, B, E and F always step by the same amount; C and D do not.
patterntermsnext minus previoussame step every time?
A3, 7, 11, 154, 4, 4yes
B20, 17, 14, 11−3, −3, −3yes
C1, 4, 9, 163, 5, 7no
D2, 4, 8, 162, 4, 8no
E5, 5, 5, 50, 0, 0yes
F1, 1.5, 2, 2.50.5, 0.5, 0.5yes
In C and D the step keeps changing: the squares grow by 3, 5, 7, and the doublings by 2, 4, 8.
What counts is only the pattern of differences. Whether the terms go up (A) or down (B), stay put (E), or are not whole numbers (F) makes no difference.
The rule: next term minus previous term is always the same number.
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 definition
An arithmetic sequence is a sequence in which each term minus the term before it is the same constant dd, the common difference.
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) →
So A (d=4d = 4), B (d=−3d = -3), E (d=0d = 0) and F (d=0.5d = 0.5) are arithmetic; C and D are not.
An arithmetic sequence is a linear function read at the whole-number inputs 1, 2, 3, …
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.
Arithmetic terms lie on straight lines: d = 4 rises, d = -3 fallsny12345648121620243, 7, 11, 15, 19: d = 4, slope 420, 17, 14, 11, 8: d = −3, slope −3Plotted against n, an arithmetic sequence is a line of slope d
The common difference is a slope

The section’s two sequences plotted as points (n, aₙ). Each step to the right adds d to the height, so the points sit on a straight line: 3, 7, 11, … rises 4 per step, and 20, 17, 14, … falls 3 per step. The common difference is the slope of that line, which is the link between aₙ = a₁ + (n − 1)d and f(x) = mx + b.

A non-arithmetic sequence such as 1, 4, 9, 16 bends away from any line.
An arithmetic sequence is a linear function read at whole-number inputs, with the common difference as its slope.
A few equal steps at the start are not proof: the pattern can break later.
1,3,5,81, 3, 5, 8 begins with steps of 22 and 22, then jumps by 33. How to check a sequence properly is the subject of identifying arithmetic sequences.
The rest of the page covers notation, identifying arithmetic sequences, finding terms, the recursive formula, arithmetic series, and the arithmetic mean.

Sequence Notation

Notation

Sequence Notation

The subscript that means position, the recurrence written in two index frames, and the two compressions — braces for the whole sequence, sigma for its sum.
ana_n
a sub n — the n-th term
The subscript is a position: a7a_7 is whatever sits in seventh place. One letter plus a counter names infinitely many values — the counter nn ranges, unlike the fixed job-title subscripts of polynomial coefficients or the two-name enumeration of quadratic roots.
CasesWhere counting starts is a convention: this page and most school texts begin at a1a_1; computer science and many analysis texts begin at a0a_0, which shifts every formula — the general term becomes a0+nda_0 + nd, not a1+(n−1)da_1 + (n-1)d.
Also writtena(n)a(n) — function notation, honest about what a sequence is (a function on the integers) but rare in print.
Do not confuseAn exponent's job. ana_n selects, ana^n multiplies — the raised/lowered distinction carries the entire meaning.
an+1=an+d  ≡  an=an−1+da_{n+1} = a_n + d \;\equiv\; a_n = a_{n-1} + d
Each term is the previous term plus d — written from either end
One rule, two index frames: stepping forward from nn, or looking back from nn. The Definition above uses the first; the Recursive Formula below uses the second. They differ by nothing but where the window sits.
CasesA recurrence is incomplete without its anchor: a1=ca_1 = c must ride along, or the rule generates no particular sequence at all — every choice of first term satisfies it.
Also writtenan+1−an=da_{n+1} - a_n = d — the difference form, stating the same rule as a property rather than a recipe.
Do not confuseTwo different rules. Shifting every index by one changes no content — but mixing frames mid-computation, reading an−1a_{n-1} in one line as the ana_n of the next, silently shifts the whole sequence by a step.
{an}\{a_n\} · ∑k=1nak\sum_{k=1}^{n} a_k
The sequence a-n; the sum of a-k, for k from one to n
Two compressions. Braces promote the terms to a single object — the whole sequence. Sigma compresses the sum SnS_n of Arithmetic Series below: the index kk starts at the lower bound, walks to the upper, and contributes one term per step.
CasesThe summation index is a dummy — ∑ak\sum a_k and ∑aj\sum a_j are the same number, and kk vanishes from the answer, exactly like the definite integral's variable of integration.
Also writtena1+a2+⋯+ana_1 + a_2 + \cdots + a_n — the dotted form the body uses; sigma is its exact abbreviation. (an)(a_n) — parentheses instead of braces, common in analysis.
Do not confuseSet braces. {an}\{a_n\} is ordered and repeats freely; a set {2,4,6}\{2, 4, 6\} is unordered with duplicates erased — same glyphs, incompatible objects, which is why analysis texts prefer (an)(a_n).
DownloadPNG2×PDF
Notation · Cheat sheet
Sequence Notation at a Glance
The marks from the notation cards across the sequence pages: how each is read and what it means. Every group links to the full card on its page.
Sequences
ana_n
a sub n
The term in place nn; the subscript is a position.
a1,a2,a3,…a_1, a_2, a_3, \ldots
a one, a two, a three, and so on
The terms listed in order; the dots need a stated rule.
{an}n=1∞\{a_n\}_{n=1}^{\infty}
the sequence a sub n
The whole sequence as one object.
Arithmetic sequencesThis page
an+1=an+da_{n+1} = a_n + d
next term is this term plus d
The recurrence; dd is the common difference, a1a_1 the start.
∑k=1nak\textstyle\sum_{k=1}^{n} a_k
sum of a sub k, k from 1 to n
Sigma: the sum SnS_n of the first nn terms.
Geometric sequences
rr
r, the common ratio
The factor from each term to the next: an=a1rn−1a_n = a_1 r^{n-1}.
SnS_n
S sub n
Sum of the first nn terms; nn counts terms.
SS
S, the infinite sum
Equals a1/(1−r)a_1 / (1 - r), only when ∣r∣<1|r| < 1.
GG
G, the geometric mean
G=abG = \sqrt{ab}, the middle value of a,G,ba, G, b.
Fibonacci sequence
FnF_n
F sub n
The nn-th Fibonacci number, counted from F1=F2=1F_1 = F_2 = 1.
ϕ,  ψ\phi, \; \psi
phi and psi
The roots of x2=x+1x^2 = x + 1, about 1.6181.618 and −0.618-0.618.
LnL_n
L sub n
Lucas numbers: the same rule, started at 1,31, 3.
Harmonic sequences
an=1/bna_n = 1/b_n
a sub n is one over b sub n
bnb_n is the arithmetic sequence underneath.
HnH_n
H sub n
The harmonic number 1+12+⋯+1n1 + \frac{1}{2} + \cdots + \frac{1}{n}.
HH
H, the harmonic mean
For two numbers, H=2aba+bH = \frac{2ab}{a + b}.
Triangular numbers
TnT_n
T sub n
The nn-th triangular number, n(n+1)2\frac{n(n+1)}{2}.
Tn=Tn−1+nT_n = T_{n-1} + n
the one before plus a row of n
The recurrence, from T1=1T_1 = 1.
(n+12)\binom{n+1}{2}
n plus one choose two
The same number, counted as pairs.
Square numbers
Sn=n2S_n = n^2
S sub n on this page
The nn-th square: one term, not a sum.
Sn=Sn−1+2n−1S_n = S_{n-1} + 2n - 1
the one before plus the next odd number
The recurrence, from S1=1S_1 = 1.
∑k=1nk2\textstyle\sum_{k=1}^{n} k^2
sum of k squared
Square each term first, then add.
Prime numbers
pp
p, a prime
The letter promises a prime.
p1,…,pkp_1, \ldots, p_k
p sub one to p sub k
A list of primes, or the primes of one number.
π(n)\pi(n)
pi of n
How many primes are at most nn.
Same letter, different job
SnS_nA sum of nn terms on the arithmetic and geometric pages; a single square on the square numbers page.
HnH_n vs HHWith a subscript, a sum of reciprocals; without one, a mean.
π(n)\pi(n) vs π\piA count of primes, not the circle constant.
{an}\{a_n\}An ordered sequence, not a set: terms keep their order and may repeat.
Sequence symbols tableSequences
Sequence Notation at a Glance·/algebra/sequencesLearn Math Class

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.)

Pitfall: Subtracting in the wrong order
Taking each term minus the next one: for 20,17,14,1120, 17, 14, 11 that gives 20−17=320 - 17 = 3, so d=3d = 3, as if the sequence rose. The difference is later minus earlier, 17−20=−317 - 20 = -3, so d=−3d = -3.

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.

Pitfall: Counting n steps instead of n − 1
Writing a20=4+20⋅6=124a_{20} = 4 + 20 \cdot 6 = 124 adds dd once too often, since a1a_1 already stands in place 11. Only 1919 steps lead to place 2020: 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.

Example
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.

Fact: One square brings infinitely many
If an arithmetic sequence of whole numbers with d>0d > 0 contains one perfect square x2x^2, it contains infinitely many. The reason: (x+d)2=x2+d(2x+d)(x + d)^2 = x^2 + d(2x + d), so (x+d)2(x + d)^2 lies a whole number of steps further along, and the same step repeats forever.

The same position test shows which numbers a sequence can hold. In 3,7,11,15,…3, 7, 11, 15, \ldots a value vv appears only when v−34\frac{v - 3}{4} is a whole number, that is, when vv leaves remainder 33 on division by 44. Every perfect square leaves remainder 00 or 11, so this sequence contains no square at all. The sequence 1,4,7,10,…1, 4, 7, 10, \ldots, by contrast, already holds 11, 44 and 1616 among its first six terms.

To find any term of your own arithmetic sequence, use the arithmetic sequence calculator.
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
Every term-finding problem comes down to the general term an=a1+(n−1)da_n = a_1 + (n - 1)d, solved for whatever is missing.

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:

Arithmetic Series
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:

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

Arithmetic Series
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:

Arithmetic Series
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:

Arithmetic Series
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.

Example
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.

Fact: Only powers of 22 resist
Every whole number except 1,2,4,8,16,…1, 2, 4, 8, 16, \ldots is a sum of two or more consecutive positive integers. Such a sum is k(2a+k−1)2\frac{k(2a + k - 1)}{2}, where one of kk and 2a+k−12a + k - 1 is odd and larger than 11, so a power of 22 can never be one.

With d=1d = 1 the formula sums any run of consecutive whole numbers. A run of kk terms starting at aa ends at a+k−1a + k - 1, so its sum is k2(2a+k−1)\frac{k}{2}(2a + k - 1). Read the other way, the formula splits a number into such runs: 15=7+8=4+5+6=1+2+3+4+515 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5, and 9=4+5=2+3+49 = 4 + 5 = 2 + 3 + 4.
Columns 2+14, 5+11, 8+8, 11+5, 14+2 all reach 16, so 2S = 5 x 16ny1234564812162021451188115142every column: 2 + 14 = 162S₅ = 5 · 16S₅ = 40Forwards (blue) plus backwards (amber): 5 columns of 16 each
Forwards plus backwards

The terms 2, 5, 8, 11, 14 stand as blue columns. The same terms in reverse order, 14, 11, 8, 5, 2, are stacked on top in amber. Each column then reaches 2 + 14 = 16, so the two copies together fill 5 columns of 16: 2S₅ = 5 · 16 = 80 and S₅ = 40. That is Sₙ = n(a₁ + aₙ)/2.

Gauss's 1 + 2 + … + 100 is the same picture with 100 columns of 101.
Any arithmetic sum equals the number of terms times the average of the first and last terms.

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.

Example
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.

Pitfall: Dividing by the number of means
Dividing 25−525 - 5 by 33 gives d=203d = \frac{20}{3} and the list 5,353,553,255, \frac{35}{3}, \frac{55}{3}, 25, with only two means. Three means make four steps: d=204=5d = \frac{20}{4} = 5.

In this kind of problem the two endpoints are fixed and only the spacing between them is unknown. Once dd is found, the inserted terms follow by repeated addition, and the last step from a+kda + kd must land exactly on bb, which checks the whole list at once.

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

Common Pitfalls

Three mistakes come up again and again with arithmetic sequences. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
DownloadPNG2×PDF
1Subtracting in the wrong order
When the consecutive differences are computed, each term is subtracted from the one before it instead of the one after it. The differences still come out equal, but with the wrong sign.
Start
For 20,17,14,1120, 17, 14, 11,
Wrong turn
Error here
20−17=3,17−14=3,14−11=320 - 17 = 3, \quad 17 - 14 = 3, \quad 14 - 11 = 3

  ⟹  d=3\implies d = 3
Right path
The common difference is d=an+1−and = a_{n+1} - a_n, later term minus earlier term.
17−20=−3,14−17=−3,17 - 20 = -3, \quad 14 - 17 = -3,

11−14=−311 - 14 = -3

  ⟹  d=−3\implies d = -3
Why it happensThe larger number is written first out of habit, and because all the differences agree, the constant-difference test passes and hides the sign error.
How to checkAdd dd to the first term; it must give the second. 20+3=2320 + 3 = 23, not 1717, while 20+(−3)=1720 + (-3) = 17. A falling sequence always has d<0d < 0.
2Counting n steps instead of n − 1
In the general term, dd is multiplied by the position nn instead of by n−1n - 1, the number of steps from the first term.
Start
With a1=4a_1 = 4 and d=6d = 6,
Wrong turn
Error here
a20=4+20⋅6=124a_{20} = 4 + 20 \cdot 6 = 124
Right path
From place 11 to place 2020 there are 1919 steps.
a20=4+(20−1)⋅6=4+114=118a_{20} = 4 + (20 - 1) \cdot 6 = 4 + 114 = 118
The same slip appears when solving for dd: from a1=3a_1 = 3 and a8=24a_8 = 24, dividing 24−3=2124 - 3 = 21 by 88 gives d=218d = \frac{21}{8} instead of 217=3\frac{21}{7} = 3.
Why it happensThe formula is remembered as "first term plus nn times dd", and the fact that a1a_1 already occupies the first place, before any dd is added, gets lost.
How to checkPut n=1n = 1 into the formula; it must return a1a_1. The wrong version gives 4+1⋅6=104 + 1 \cdot 6 = 10, not 44, while 4+0⋅6=44 + 0 \cdot 6 = 4.
3Dividing by the number of means
When kk arithmetic means are inserted between aa and bb, the gap b−ab - a is split into kk equal parts instead of k+1k + 1.
Start
33 means between 55 and 2525,
Wrong turn
Error here
d=25−53=203d = \frac{25 - 5}{3} = \frac{20}{3}
Which gives the list 5,353,553,255, \frac{35}{3}, \frac{55}{3}, 25, with only two values between the endpoints.
Right path
33 means and 22 endpoints make 55 terms, and 55 terms have 44 steps between them.
d=25−53+1=5d = \frac{25 - 5}{3 + 1} = 5
The sequence is 5,10,15,20,255, 10, 15, 20, 25.
Why it happensThe number of values to insert and the number of steps between neighbouring values are easy to confuse; there is always one more step than there are inserted values.
How to checkCount the values strictly between the endpoints in the finished list; there must be exactly kk. Equivalently, a+(k+1)da + (k + 1)d must equal bb: 5+4⋅5=255 + 4 \cdot 5 = 25.
Common pitfalls with arithmetic sequences·/algebra/sequences/arithmeticLearn Math Class
All three slips miscount or misread the steps between terms. The difference is always a later term minus an earlier one, place nn is n−1n - 1 steps after place 11, and kk means between two endpoints make k+1k + 1 steps. Putting n=1n = 1 into a formula, adding dd to a term to see whether the next one appears, or counting the values in the finished list catches each slip in a line.

Arithmetic Sequences FAQ

Should a sequence start at a1a_1 or a0a_0?

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Both conventions are in use, and the choice changes every formula. This page and most school texts begin at a1a_1, giving the general term a1+(n−1)da_1 + (n-1)d. Computer science and many analysis texts begin at a0a_0, where the same sequence has general term a0+nda_0 + nd. Neither is wrong, but mixing them shifts your answer by one full step.Read more →

Is {an}\{a_n\} a set?

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No, despite the braces. A sequence is ordered and repeats freely, so {1,1,2,1}\{1, 1, 2, 1\} is a perfectly good sequence. A set is unordered and erases duplicates, so the same braces there would collapse to {1,2}\{1, 2\}. The glyphs collide but the objects are incompatible, which is why analysis texts often write (an)(a_n) instead.Read more →

Why isn't an=an−1+da_n = a_{n-1} + d enough to define a sequence?

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Because a recurrence without an anchor describes infinitely many sequences at once. The rule says each term exceeds the previous one by dd, but says nothing about where to begin, so every choice of first term satisfies it equally. Pairing it with an initial value such as a1=3a_1 = 3 is what pins down one particular sequence.Read more →

Can an arithmetic sequence contain perfect squares?

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Yes, many do, and a sequence of whole numbers with d>0d > 0 that holds one square holds infinitely many. If x2x^2 is a term, then (x+d)2=x2+d(2x+d)(x + d)^2 = x^2 + d(2x + d) lies a whole number of steps further on, and the same step repeats forever. Some sequences hold none at all: every term of 3,7,11,15,…3, 7, 11, 15, \ldots leaves remainder 33 on division by 44, while every square leaves 00 or 11.Read more →

Which numbers can be written as a sum of consecutive positive integers?

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Every positive whole number except the powers of 22, that is 1,2,4,8,16,…1, 2, 4, 8, 16, \ldots. A run of k≥2k \geq 2 consecutive positive integers starting at aa adds up to k(2a+k−1)2\frac{k(2a + k - 1)}{2}, and one of the factors kk and 2a+k−12a + k - 1 is odd and larger than 11, which a power of 22 never has. Other numbers often split in several ways: 15=7+8=4+5+6=1+2+3+4+515 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5.Read more →

Do the first few terms prove that a sequence is arithmetic?

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No. A finite list can only rule the pattern out: one pair of unequal differences disqualifies it, but equal differences so far confirm nothing. The list 2,4,62, 4, 6 could go on as 8,10,…8, 10, \ldots or as 10,16,…10, 16, \ldots, and 1,3,5,81, 3, 5, 8 begins with two steps of 22 before jumping by 33. Only the rule that generates the sequence settles its type.Read more →