The Arithmetic Series Calculator computes the sum of the first n terms of an arithmetic progression from the first term, common difference, and number of terms. Essential for algebra, financial mathematics, and any problem requiring the sum of a regularly spaced progression.
1,010
98
50.5
23
48
1,010
98
50.5
23
48
An arithmetic series is the sum of an arithmetic sequence — you are not just listing the terms but adding them all up. The calculator for arithmetic series computes Sₙ instantly from any combination of first term, last term, common difference, and number of terms, making it easy to solve both forward problems (given a₁, d, n, find the sum) and backward problems (given the sum and some terms, find the missing parameters).
The sum of the first n terms of an arithmetic series with first term a₁ and common difference d:
Sₙ = n/2 × (2a₁ + (n−1)d) = n/2 × (a₁ + aₙ)
where aₙ = a₁ + (n−1)d is the nth (last) term. Both forms are equivalent — use whichever variables you have available. For the series 2 + 5 + 8 + ... + 50 (a₁ = 2, d = 3, aₙ = 50): n = (50 − 2)/3 + 1 = 17 terms; S₁₇ = 17/2 × (2 + 50) = 17/2 × 52 = 442. Alternatively: S₁₇ = 17/2 × (2×2 + 16×3) = 17/2 × 52 = 442 ✓. Use this online calculator for any arithmetic series. The arithmetic sequence calculator handles the term-by-term analysis when individual terms are needed.
The most famous arithmetic series is the sum of consecutive integers 1 + 2 + 3 + ... + n = n(n+1)/2. This is the arithmetic series with a₁ = 1, d = 1: Sₙ = n/2 × (1 + n) = n(n+1)/2. Legend holds that 8-year-old Carl Friedrich Gauss surprised his teacher by instantly computing 1 + 2 + ... + 100 = 5,050 by pairing 1 with 100, 2 with 99, etc. — each pair summing to 101, with 50 pairs total: 50 × 101 = 5,050. This insight — pairing the first and last terms — is exactly the derivation of the arithmetic series formula.
Arithmetic series arise naturally in financial problems involving linear (not compounding) growth:
The geometric series calculator handles the multiplicative counterpart for compound growth scenarios. The sequences and series calculators provide the complete mathematical analysis toolkit.
Backward problems — given Sₙ, find a₁, d, or n — require solving the series formula for the unknown variable. Finding n from Sₙ, a₁, and d involves the quadratic formula: n² × d + n × (2a₁ − d) − 2Sₙ = 0, solved by: n = [−(2a₁ − d) + √((2a₁ − d)² + 8dSₙ)] / (2d). For Sₙ = 120, a₁ = 3, d = 2: n² + 2n − 120 = 0 → (n+12)(n−10) = 0 → n = 10 terms. This type of reverse calculation appears in optimization problems and in fitting arithmetic models to known totals.
Step 1: General term formula. The $$n$$-th term of an arithmetic sequence with first term $$a_1$$ and common difference $$d$$ is:
$$a_n = a_1 + (n - 1)d$$
Step 2: Sum formula. The sum of the first $$n$$ terms is derived by writing the series forward and backward, then adding corresponding pairs:
$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d)$$
Each pair sums to $$a_1 + a_n$$, and there are $$n/2$$ such pairs (or $$n$$ half-pairs), giving the formula.
Step 3: Arithmetic mean.
$$\bar{a} = \frac{S_n}{n} = \frac{a_1 + a_n}{2}$$
The arithmetic mean of an arithmetic series always equals the average of its first and last terms, which is also the median of the series.
Step 4: Sample terms. The 5th and 10th terms are computed as $$a_5 = a_1 + 4d$$ and $$a_{10} = a_1 + 9d$$, providing checkpoints to verify the progression.
Sum Sₙ is the total of all $$n$$ terms. For large $$n$$, the sum grows quadratically because the terms themselves grow linearly.
Last Term aₙ reveals how far the sequence has progressed from $$a_1$$. If $$d$$ is negative, $$a_n$$ will be less than $$a_1$$.
Arithmetic Mean is the average value of the terms. In an arithmetic series, this always equals the midpoint between the first and last terms, regardless of how many terms there are.
5th and 10th Terms are sample values that help you verify the sequence is progressing as expected. They are especially useful for checking your parameters before computing large sums.
Inputs
Results
S₁₀₀ = 100/2 × (1 + 100) = 50 × 101 = 5050. This is Gauss's classic result. The arithmetic mean is (1 + 100)/2 = 50.5, confirming the pairing argument.
Inputs
Results
a₁₅ = 50 + 14×(−3) = 50 − 42 = 8. S₁₅ = 15/2 × (50 + 8) = 7.5 × 58 = 435. The mean is 435/15 = 29, which equals (50 + 8)/2 = 29.
An arithmetic sequence is an ordered list of numbers with a constant difference between consecutive terms (e.g., 2, 5, 8, 11, …). An arithmetic series is the sum of those terms (e.g., 2 + 5 + 8 + 11 + … = S). The sequence gives individual terms; the series gives accumulated totals.
Yes. When $$d = 0$$, every term equals $$a_1$$, and the sum is simply $$S_n = n \cdot a_1$$. This is a trivial but valid arithmetic series.
Yes. A negative $$d$$ produces a decreasing arithmetic sequence. The sum formula works identically — the terms get smaller, and eventually become negative if the sequence runs long enough. The sum may be positive, zero, or negative depending on the parameters.
No. Unless $$a_1 = 0$$ and $$d = 0$$ (the trivial zero series), the terms of an arithmetic sequence do not approach zero, so by the divergence test, the infinite arithmetic series always diverges. Only the finite partial sum $$S_n$$ is meaningful.
The arithmetic series sum $$S_n = \frac{n}{2}(a_1 + a_n)$$ has the same structure as the trapezoidal rule for numerical integration: average the first and last values, then multiply by the width. This is not a coincidence — a linear function sampled at equal intervals produces an arithmetic sequence, and the trapezoidal rule is exact for linear functions.
Rearrange the general term formula: $$d = \frac{a_n - a_1}{n - 1}$$. For example, if $$a_1 = 3$$, $$a_{20} = 60$$, then $$d = (60 - 3)/19 = 3$$.
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