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Arithmetic Sequence Calculator

How to use Arithmetic Sequence Calculator
  1. Set First term a₁ and Common difference d in the Parameters row and click Apply; every tab recomputes with the new rule. Learn more about the parameters
  2. Pick a mode with the five tabs across the top of the card: List first N, List a – b, Test a number, Lookup by index or Solver; the ? beside each tab describes it. Learn more about the modes
  3. In List first N, type how many terms to show in Show first and click Update; the cards give the count, the largest term and the sum, and the tiles list every term. Learn more about listing terms
  4. In List a – b, type the From index and the To index and click List range; a red box explains a bad range. Learn more about ranges
  5. In Test a number, type a number in Test number and click Check; a green box names the index, a red box says it is not a term. Learn more about the test
  6. In Lookup by index, type the position in Index n and click Get term. Learn more about lookups
  7. In Solver, pick Find term (aₙ) or Sum Sₙ, fill three boxes, leave the one marked solve empty and click Solve; move the radio to change the unknown. Learn more about the solver
  8. The panel on the right explains the current mode and walks through the arithmetic with your own numbers; it changes with every tab and click. Learn more about the panel





Setting the Parameters

DemoTwo new parameter pairs, then the opening pair again
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The card opens with the Sequence drop-down set to Arithmetic sequence; the other entry switches the card to the Geometric sequence, which has its own page, Geometric sequence calculator. Under it the Parameters row holds the two numbers that fix the sequence, First term a₁ and Common difference d, opening with a1=2a_1 = 2 and d=3d = 3. Type new values and click Apply or press Enter; every tab recomputes at once with the rule an=a1+(n−1)⋅da_n = a_1 + (n - 1) \cdot d. With a1=5a_1 = 5 and d=2d = 2 the first eight terms run 5,7,…,195, 7, \ldots, 19 with the sum 9696; with a1=10a_1 = 10 and d=−3d = -3 they fall from 1010 to −11-11 and the sum is −4-4.

Both boxes take any decimal, negative values included; a blank or non-numeric box is refused with "Enter a number in every parameter box." and the tabs keep their last values. Which two numbers define the sequence is explained in what is an arithmetic sequence.

Listing the First N Terms

DemoThree counts in List first N
Step 0 of 3
The explorer opens on the List first N tab, one of five tabs across the top of the card; the small ? beside each tab shows what the mode does. Type how many terms you want in the Show first box and click Update or press Enter. Three cards answer at once: Showing gives the count, Largest the last term with its index, and Sum the total of the listed terms, SN=N2(a1+aN)S_N = \frac{N}{2}(a_1 + a_N). Under the cards every term is laid out as a tile with its index above its value.

With the opening parameters the box accepts 11 to 500500: 88 gives a8=23a_8 = 23 with the sum 100100, 1010 gives a10=29a_{10} = 29 with the sum 155155, and 600600 is capped at 500500 terms, ending at a500=1,499a_{500} = 1,499 with the sum 375,250375,250; a single term gives a1=2a_1 = 2. Large values are printed with thousands separators. A count of 0, a decimal or an empty box is refused with "Enter a positive integer." There is no Reset button: type a new count and click Update again. The panel on the right, List first N, states the rule the tiles come from, and its walkthrough repeats the last term and the sum with your own NN. The rule is explained in the n-th term formula and the sum in the sum formula and gauss's trick.

Listing a Range of Indexes

DemoThe range tab, a new range, two bad ranges
Step 0 of 4
Click the List a – b tab to list the terms between two positions instead of starting from 11. Type the From index and the To index and click List range or press Enter. The cards give the Count b−a+1b - a + 1, the Smallest term aaa_a and the Largest term aba_b, and the tiles below list every term in between with its index. The tab opens with 55 to 1212, 8 terms from a5=14a_{5} = 14 to a12=35a_{12} = 35; indexes 1010 to 2020 give 11 terms from 2929 to 5959.

Both boxes must hold positive integers, the first no larger than the second, and the range may hold at most 500500 terms; otherwise a red box explains: "Enter two positive integer indexes", "From-index must be less than or equal to to-index", or "Range too wide — cap is 500 terms". The panel on the right, List a – b, explains the mode, and its walkthrough names the smallest and largest term of your range. The rule each tile uses is explained in the n-th term formula.

Testing a Number

DemoThe test tab, a non-member, a member, a decimal
Step 0 of 4
Click the Test a number tab to ask whether a given number is a term of the sequence. Type it in the Test number box and click Check or press Enter. A green box with a tick answers yes and names the index, as for the opening value 4747, which is a16a_{16}; a red box with a cross answers no, as for 4949, and the walkthrough on the right then shows the two nearest terms, a16=47a_{16} = 47 and a17=50a_{17} = 50. A large value is handled the same way: 2,9992,999 is a1000a_{1000}.

The box takes any number, decimals and negatives included, since the parameters may be decimals too; an empty box is refused with "Enter a number". The test is exact: the explorer computes k=(m−a1)/dk = (m - a_1) / d and accepts mm only when kk is a whole number that is not negative; the index is then k+1k + 1. The panel on the right, Test a number, states this rule, and its walkthrough carries out the arithmetic for your number. The formula behind it is in the n-th term formula.

Looking Up a Term by Index

DemoThe lookup tab, two indexes, n = 0
Step 0 of 4
Click the Lookup by index tab to compute one term directly. Type the position in the Index n box and click Get term or press Enter. Two cards answer: Index repeats nn and Value gives ana_n with its index as a subscript. The tab opens with n=20n = 20, which gives a20=59a_{20} = 59; n=100n = 100 gives 299299 and n=1000n = 1000 gives 2,9992,999. The box takes positive integers only; 00 is refused with "Enter a positive integer index".

The panel on the right has two parts in every mode. How it works names the rule the mode uses, here an=a1+(n−1)⋅da_n = a_1 + (n - 1) \cdot d, and links to the section of this page that explains it; Walkthrough applies the rule to the numbers you typed, so for the lookup it shows the arithmetic with your nn and the current parameters filled in. Both change as soon as you switch tabs or click a button. The rule and where it comes from are in the n-th term formula.

Using the Solver Tab

DemoThe solver tab: a term, an unknown moved, a sum
Step 0 of 4
Click the Solver tab, the fifth tab, to work the formulas backwards. Two buttons choose the relation: Find term (aₙ) uses an=a1+(n−1)⋅da_n = a_1 + (n - 1) \cdot d and Sum Sₙ uses Sn=n2(2a1+(n−1)⋅d)S_n = \frac{n}{2}(2a_1 + (n - 1) \cdot d). Each relation has four boxes; the one marked solve with a ? is the unknown and the other three must be filled. The unknown opens as ana_n or SnS_n; click the solve radio beside another box to make that the unknown. Click Solve or press Enter and the answer appears under the grid, while the walkthrough on the right shows the substitution.

With a1=2a_1 = 2, d=3d = 3 and n=20n = 20 the term is a20=59a_{20} = 59; with an=299a_n = 299 and the unknown moved to nn the solver returns n=100n = 100; with n=10n = 10 the sum is S10=155S_{10} = 155. Switching the relation clears the boxes, and clicking Solve with a box empty gives the message "Fill all fields except the one being solved for". Solving SnS_n for nn solves a quadratic and reports "No positive solution for n" when none fits. The solver boxes are independent of the Parameters row. Every relation is explained in using the solver — five quantities, four relations.

What is an Arithmetic Sequence?

An arithmetic sequence (also called an arithmetic progression, AP) is a sequence where each term is obtained by adding a constant to the previous term. The constant is called the common difference, denoted dd.

a1,    a1+d,    a1+2d,    a1+3d,    …a_1, \;\; a_1 + d, \;\; a_1 + 2d, \;\; a_1 + 3d, \;\; \ldots


Examples.

• a1=2a_1 = 2, d=3d = 3: gives 2,5,8,11,14,17,20,23,…2, 5, 8, 11, 14, 17, 20, 23, \ldots
• a1=10a_1 = 10, d=−3d = -3: gives 10,7,4,1,−2,−5,−8,…10, 7, 4, 1, -2, -5, -8, \ldots
• a1=0a_1 = 0, d=1d = 1: gives 0,1,2,3,4,…0, 1, 2, 3, 4, \ldots — the natural numbers.
• a1=1a_1 = 1, d=2d = 2: gives 1,3,5,7,9,…1, 3, 5, 7, 9, \ldots — the odd numbers.
• a1=100a_1 = 100, d=0d = 0: gives 100,100,100,…100, 100, 100, \ldots — the constant sequence.

Two key formulas.

• The nn-th term: an=a1+(n−1)da_n = a_1 + (n - 1) d.
• The sum of the first nn terms: Sn=n2(a1+an)=n2(2a1+(n−1)d)S_n = \dfrac{n}{2}(a_1 + a_n) = \dfrac{n}{2}(2 a_1 + (n - 1) d).

Parametric nature. Unlike triangular or square numbers (where the sequence is fixed), an arithmetic sequence depends on two parameters: a1a_1 and dd. Different choices give different sequences. This explorer lets you set both parameters and explore the resulting sequence.

For deeper coverage see arithmetic sequence, arithmetic progression, and series.

The definition and the notation are explained in Arithmetic sequences: definition and Sequence notation; the overview page compares the main types in Arithmetic sequences.

Finding terms works the formula through more examples, and Recursive formula gives the step-by-step form an=an−1+da_n = a_{n-1} + d.

Arithmetic series proves the sum formula and applies it; Explicit and recursive definitions places sums next to the term rules.

Each relation the solver uses is derived in Finding terms and Arithmetic series.

The mean property is the subject of Arithmetic mean; the recurrence is in Recursive formula.

The n-th Term Formula

Derivation. Starting from a1a_1, the next term is a1+da_1 + d, then a1+2da_1 + 2d, then a1+3da_1 + 3d. After n−1n - 1 additions, the nn-th term is

an=a1+(n−1)da_n = a_1 + (n - 1) d


Inverting the formula. Given ana_n, a1a_1, and dd, solve for nn:

n=an−a1d+1n = \frac{a_n - a_1}{d} + 1


This works only when d≠0d \neq 0; if d=0d = 0, every term equals a1a_1 and nn is undetermined.

Examples.

• Sequence 5,8,11,14,…5, 8, 11, 14, \ldots: a1=5a_1 = 5, d=3d = 3. The 20th term is a20=5+19⋅3=62a_{20} = 5 + 19 \cdot 3 = 62.
• Sequence 100,97,94,…100, 97, 94, \ldots: a1=100a_1 = 100, d=−3d = -3. The 10th term is a10=100+9⋅(−3)=73a_{10} = 100 + 9 \cdot (-3) = 73.
• Membership test: is 47 in the sequence 2,5,8,11,…2, 5, 8, 11, \ldots? Solve 47=2+(n−1)⋅3  ⟹  n=1647 = 2 + (n-1) \cdot 3 \implies n = 16. Since nn is a positive integer, yes — 47=a1647 = a_{16}.
• Non-membership: is 50 in the sequence 2,5,8,11,…2, 5, 8, 11, \ldots? Solve 50=2+(n−1)⋅3  ⟹  n=1750 = 2 + (n-1) \cdot 3 \implies n = 17. Wait, 1717 is an integer — so 5050 IS in the sequence. Let me try 5151: 51=2+(n−1)⋅3  ⟹  n=17.3351 = 2 + (n-1) \cdot 3 \implies n = 17.33, not an integer. So 5151 is not in the sequence.

Two-parameter recovery. Given any two terms ama_m and ana_n with m≠nm \neq n, both a1a_1 and dd can be recovered: d=(an−am)/(n−m)d = (a_n - a_m)/(n - m) and a1=am−(m−1)da_1 = a_m - (m - 1) d.

The Sum Formula and Gauss's Trick

The sum formula. The sum of the first nn terms of an arithmetic sequence is

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


Gauss's trick. Write the sum forward and backward:

Sn=a1+a2+a3+⋯+anSn=an+an−1+an−2+⋯+a1\begin{aligned} S_n &= a_1 + a_2 + a_3 + \cdots + a_n \\ S_n &= a_n + a_{n-1} + a_{n-2} + \cdots + a_1 \end{aligned}


Adding column by column: each pair (ak+an−k+1)(a_k + a_{n - k + 1}) has the same value, because in an arithmetic sequence the sum of equally-spaced-from-the-ends terms is constant. Specifically, ak+an−k+1=a1+ana_k + a_{n - k + 1} = a_1 + a_n for every kk. Summing nn such pairs:

2Sn=n(a1+an)  ⟹  Sn=n(a1+an)22 S_n = n (a_1 + a_n) \implies S_n = \frac{n (a_1 + a_n)}{2}


The Gauss schoolboy story. Legend says young Carl Friedrich Gauss was asked to sum 1+2+⋯+1001 + 2 + \cdots + 100. He pictured the trick instantly: pair up 1+100,2+99,…,50+511 + 100, 2 + 99, \ldots, 50 + 51, getting fifty pairs of 101, total 50505050. Whether the story is literal or apocryphal, the technique works for any arithmetic sequence.

Average-of-endpoints interpretation. Sn=n⋅a‾S_n = n \cdot \overline{a} where a‾=(a1+an)/2\overline{a} = (a_1 + a_n)/2 is the average of the first and last terms — which is also the arithmetic mean of the entire sequence (a special property of arithmetic sequences).

Solving for nn. Given SnS_n, a1a_1, and dd, the equation Sn=(n/2)(2a1+(n−1)d)S_n = (n/2)(2 a_1 + (n - 1) d) is quadratic in nn. Solving:

dn2+(2a1−d)n−2Sn=0d n^2 + (2 a_1 - d) n - 2 S_n = 0


The quadratic formula gives nn. There can be two positive solutions; the smaller is usually the intended one.

Using the Solver — Five Quantities, Four Relations

The arithmetic-sequence explorer includes a solver mode that exploits the relationships between the five quantities a1,d,n,an,Sna_1, d, n, a_n, S_n.

Two solver sub-modes.

• Find term (ana_n) — uses the relation an=a1+(n−1)da_n = a_1 + (n - 1) d. Fill any three of {a1,d,n,an}\{a_1, d, n, a_n\} and the solver computes the fourth.
• Sum SnS_n — uses the relation Sn=(n/2)(2a1+(n−1)d)S_n = (n/2)(2 a_1 + (n - 1) d). Fill any three of {a1,d,n,Sn}\{a_1, d, n, S_n\} and the solver computes the fourth.

Solving for ana_n. Direct substitution: an=a1+(n−1)da_n = a_1 + (n - 1) d.

Solving for a1a_1. Rearrange: a1=an−(n−1)da_1 = a_n - (n - 1) d (term sub-mode) or a1=Sn/n−(n−1)d/2a_1 = S_n/n - (n - 1) d/2 (sum sub-mode).

Solving for dd. d=(an−a1)/(n−1)d = (a_n - a_1)/(n - 1) (term sub-mode) or d=2(Sn−na1)/(n(n−1))d = 2(S_n - n a_1)/(n(n - 1)) (sum sub-mode). Both undefined when n=1n = 1.

Solving for nn. n=(an−a1)/d+1n = (a_n - a_1)/d + 1 (term sub-mode, undefined when d=0d = 0). For sum sub-mode, solve the quadratic dn2+(2a1−d)n−2Sn=0d n^2 + (2 a_1 - d) n - 2 S_n = 0; there may be two positive solutions, and the smaller positive integer is usually the answer.

Edge cases the solver handles.

• d=0d = 0: the sequence is constant; nn is undetermined for the term sub-mode (any nn gives the same value) and reduces to Sn/a1S_n / a_1 for the sum sub-mode.
• n=1n = 1: dd cannot be recovered from a single term.
• Negative or fractional nn in the quadratic: rejected (no real positive solution).

Practical use. The solver is helpful for word problems: given partial information about an arithmetic sequence, recover the missing quantity. Example: "The 5th term of an arithmetic sequence is 17 and the 10th term is 32. Find the 50th term." Solve for a1a_1 and dd from the two equations, then compute a50a_{50}.

Properties and Identities

• Recurrence: an=an−1+da_n = a_{n-1} + d with given a1a_1. The defining recurrence of any arithmetic sequence.
• Term in terms of any other term: an=am+(n−m)da_n = a_m + (n - m) d. The difference of two terms equals the index gap times dd.
• Symmetry of sums: ak+an−k+1=a1+ana_k + a_{n - k + 1} = a_1 + a_n for any kk from 1 to nn. The pairing property behind Gauss's trick.
• Arithmetic mean is the middle term: the average of nn consecutive terms of an arithmetic sequence equals the middle term (if nn is odd) or the average of the two middle terms (if nn is even). Always equals (a1+an)/2(a_1 + a_n)/2.
• Sum identity: Sn=na‾S_n = n \overline{a}, the count times the average.
• Sum of squares: not a simple closed form unless reformulated. ∑ak2\sum a_k^2 involves ∑k2\sum k^2 and is degree 3 in nn.
• Difference of sums: Sn−Sn−1=anS_{n} - S_{n-1} = a_n, the partial sums recover the original sequence.
• Common difference from sum: Sn−2Sn−1+Sn−2=dS_n - 2 S_{n-1} + S_{n-2} = d (the second difference of partial sums is the common difference).

Common Applications

• Linear pricing and depreciation. A car loses a fixed amount per year — its value follows an arithmetic sequence. Pricing schedules, simple interest, and linear cost models all use arithmetic sequences.
• Counting in lattices. Lattice points along a line with rational slope form an arithmetic sequence in one of the coordinates.
• Physics: uniform motion and constant-acceleration sums. Position under constant velocity follows an arithmetic sequence at equal time intervals. Velocity under constant acceleration does the same.
• Calendars and timing. Every 4th year is a leap year (modulo century rules). Recurring schedules form arithmetic progressions.
• Tournament round counts. Round-robin and elimination tournaments often have round counts following arithmetic patterns.
• Stack and pile counts. Arrangements of objects with constant-difference layer counts (rare in nature, common in puzzles).
• Simple interest. Account balance grows by PrP r each year, an arithmetic sequence with a1=Pa_1 = P and d=Prd = P r. (Compound interest is geometric.)
• Solving word problems. Many classical algebra problems reduce to arithmetic sequences: "Anna's salary increases by \3,000 each year. After 10 years she earns \70,000. What was her starting salary?" — find a1a_1 given a11=70000a_{11} = 70000 and d=3000d = 3000.

Identifying arithmetic sequences shows how to spot a constant difference in data; Arithmetic sequences at a glance collects the formulas used here.

Common Mistakes

• Off-by-one in the nn-th term formula. an=a1+(n−1)da_n = a_1 + (n - 1) d, not a1+nda_1 + n d. The 1st term is a1a_1 (zero dd's added), the 2nd term is a1+da_1 + d (one dd added). Many wrong answers come from forgetting the −1-1.

• Wrong sum formula. Sn=n(a1+an)/2S_n = n(a_1 + a_n)/2, or equivalently Sn=n(2a1+(n−1)d)/2S_n = n(2 a_1 + (n - 1) d)/2. Common errors: using na1n a_1 (missing the ana_n contribution), or (a1+an)/2(a_1 + a_n)/2 (missing the multiplication by nn).

• Negative common difference confusion. When d<0d < 0, the sequence decreases. All formulas still work; just be careful with signs. The sum is still n(a1+an)/2n(a_1 + a_n)/2, but an<a1a_n < a_1 now.

• Confusing arithmetic and geometric. Arithmetic adds dd; geometric multiplies by rr. "Doubling each step" is geometric (r=2r = 2), not arithmetic.

• Index conventions. Some sources start at a0a_0 instead of a1a_1, giving an=a0+nda_n = a_0 + n d. This explorer uses a1a_1 as the first term.

• Trying to use the formula for non-integer nn. an=a1+(n−1)da_n = a_1 + (n - 1) d is mathematically defined for any real nn, but only integer nn corresponds to actual sequence positions.

• Forgetting the second positive root in sum-for-nn quadratic. The quadratic dn2+(2a1−d)n−2Sn=0d n^2 + (2 a_1 - d) n - 2 S_n = 0 can have two positive solutions. The smaller (usually integer) one is the intended answer; the other might be an extraneous positive root or a valid second possibility for specific SnS_n values.

• Treating d=0d = 0 as a special exception incorrectly. When d=0d = 0, the sequence is constant, an=a1a_n = a_1 for all nn, and Sn=na1S_n = n a_1. Many formulas still work; just verify before applying.

The one-page summary Arithmetic sequences at a glance is a quick check against these slips; the definition is in Arithmetic sequences: definition.