Algebra Calculators

Arithmetic Sequence Calculator

Free arithmetic sequence calculator to find nth term, sum of series, common difference, first term & number of terms. Step-by-step solutions with formulas.

Arithmetic Sequence Calculator

Find nth Term, Sum of Series, Common Difference and More

Published: November 15, 2025 | Updated: February 11, 2026

By: RevisionTown Mathematics Team

An arithmetic sequence (also known as an arithmetic progression or AP) is one of the most fundamental patterns in mathematics. Every term after the first is formed by adding a fixed value, called the common difference, to the preceding term. This free arithmetic sequence calculator lets you find the nth term, compute the sum of an arithmetic series, determine the common difference, recover the first term, or count the number of terms, all with detailed step-by-step solutions.

Arithmetic progressions appear throughout algebra, number theory, financial planning, physics, and computer science. Whether you are preparing for SAT, ACT, GRE, IB Mathematics, A-Level, GCSE, or any standardised examination, a solid command of arithmetic sequences is essential. The calculator below handles positive, negative, and fractional values, and each result includes the exact formula substitution so you can learn while you solve.

Arithmetic Sequence Calculator

Result

What Is an Arithmetic Sequence?

An arithmetic sequence is an ordered list of numbers in which the difference between every pair of consecutive terms remains constant. That constant is called the common difference and is denoted by d. When d is positive the sequence increases; when it is negative the sequence decreases; and when d equals zero every term is identical.

Formally, a sequence \( a_1, a_2, a_3, \ldots \) is arithmetic if and only if

\[ a_{n+1} - a_n = d \quad \text{for every } n \geq 1 \]

Consider the sequence 4, 9, 14, 19, 24. Subtracting any term from its successor gives 5 each time, so d = 5 and the sequence is arithmetic. In contrast, 2, 4, 8, 16 is not arithmetic because the differences (2, 4, 8) are not constant; that pattern is geometric.

The concept traces back to ancient Greek mathematicians. Carl Friedrich Gauss famously exploited the structure of arithmetic series as a schoolboy when he instantly summed the integers from 1 to 100 by pairing the first and last terms. That insight underpins the sum formula used to this day.

Quick identification test

Compute the differences between consecutive terms. If every difference equals the same value, the sequence is arithmetic and that value is d.

Essential Arithmetic Sequence Formulas

1. Nth Term Formula (Explicit / General Term)

The explicit formula gives any term directly without computing all preceding terms:

\[ a_n = a_1 + (n - 1)\,d \]

Variables: \(a_n\) = value of the nth term, \(a_1\) = first term, \(n\) = term position (positive integer), \(d\) = common difference.

Worked example: Find the 25th term of 6, 11, 16, 21, ...

Here \(a_1 = 6\), \(d = 5\), \(n = 25\).

\(a_{25} = 6 + (25 - 1)(5) = 6 + 120 = \mathbf{126}\)

2. Sum of an Arithmetic Series

The sum of the first n terms of an arithmetic sequence is called an arithmetic series. Two equivalent formulas exist:

\[ S_n = \frac{n}{2}\bigl[2a_1 + (n-1)d\bigr] \]
\[ S_n = \frac{n}{2}(a_1 + a_n) \]

The second form is convenient when you already know the last term \(a_n\). Both yield the same result.

Worked example: Sum of the first 30 terms of 4, 7, 10, 13, ...

\(a_1 = 4,\ d = 3,\ n = 30\).

\(S_{30} = \frac{30}{2}[2(4) + (29)(3)] = 15 \times [8 + 87] = 15 \times 95 = \mathbf{1425}\)

3. Common Difference Formula

If you know two terms and their positions you can isolate d:

\[ d = \frac{a_n - a_1}{n - 1} \]

Worked example: Given \(a_1 = 5\) and \(a_{20} = 62\), find d.

\(d = \frac{62 - 5}{20 - 1} = \frac{57}{19} = \mathbf{3}\)

4. First Term Formula

Rearranging the nth-term formula to solve for \(a_1\):

\[ a_1 = a_n - (n - 1)\,d \]

Worked example: The 15th term of an AP is 72 and d = 5. Find the first term.

\(a_1 = 72 - (15 - 1)(5) = 72 - 70 = \mathbf{2}\)

5. Number of Terms Formula

To find how many terms lie between the first and last term (inclusive):

\[ n = \frac{a_n - a_1}{d} + 1 \]

Worked example: How many terms are in the sequence 7, 12, 17, ..., 152?

\(n = \frac{152 - 7}{5} + 1 = \frac{145}{5} + 1 = 29 + 1 = \mathbf{30}\)

6. Recursive Formula

The recursive definition builds each term from the one before it:

\[ a_n = a_{n-1} + d, \quad a_1 \text{ given} \]

This form is especially useful in programming and spreadsheet models where you iterate through terms one at a time.

How to Use the Arithmetic Progression Calculator

Step 1: Choose Your Calculation Type

Open the dropdown menu at the top of the calculator and select what you want to find. Five modes are available: nth term, sum of series, common difference, first term, and number of terms.

Step 2: Enter the Known Values

The input fields update automatically to match the selected mode. Enter each value precisely. Decimals and negative numbers are fully supported. For instance, a decreasing sequence like 50, 44, 38 has \(d = -6\).

Step 3: Click Calculate and Review

Press the blue Calculate button. The result panel displays the answer together with every substitution step so you can trace the logic. If an input is missing or invalid, a clear error message appears instead.

Step 4: Verify Using an Alternative Formula

Most arithmetic sequence problems can be checked with a second method. For example, after finding \(a_{20}\) with the explicit formula, you can verify the sum by computing \(S_{20} = \frac{20}{2}(a_1 + a_{20})\) and confirming it matches \(S_{20}\) obtained from the other sum formula.

Arithmetic Sequence Examples with Detailed Solutions

Example 1: Finding the 52nd Term

Problem: Find the 52nd term of the arithmetic sequence 3, 8, 13, 18, 23, ...

Solution:

Identify the known values: \(a_1 = 3\), \(d = 8 - 3 = 5\), \(n = 52\).

Apply the nth term formula:

\(a_{52} = 3 + (52 - 1) \times 5 = 3 + 51 \times 5 = 3 + 255 = \mathbf{258}\)

Example 2: Sum of First 30 Terms

Problem: Calculate the sum of the first 30 terms of 4, 7, 10, 13, 16, ...

Solution:

\(a_1 = 4\), \(d = 3\), \(n = 30\).

\(S_{30} = \frac{30}{2}[2(4) + (30-1)(3)] = 15[8 + 87] = 15 \times 95 = \mathbf{1425}\)

Verification: First find \(a_{30} = 4 + 29 \times 3 = 91\). Then \(S_{30} = \frac{30}{2}(4 + 91) = 15 \times 95 = 1425\). Confirmed.

Example 3: Finding the Common Difference

Problem: In an arithmetic sequence the 1st term is 12 and the 15th term is 82. Find d.

Solution:

\(d = \frac{a_{15} - a_1}{15 - 1} = \frac{82 - 12}{14} = \frac{70}{14} = \mathbf{5}\)

Example 4: Negative Common Difference

Problem: An arithmetic sequence begins 100, 93, 86, 79, ... Which term first becomes negative?

Solution:

\(a_1 = 100\), \(d = -7\). Set \(a_n < 0\):

\(100 + (n-1)(-7) < 0 \implies 100 - 7n + 7 < 0 \implies 107 < 7n \implies n> 15.28\)

The 16th term is the first negative term: \(a_{16} = 100 + 15(-7) = 100 - 105 = -5\).

The Arithmetic Mean in Sequences

The arithmetic mean of two numbers \(a\) and \(b\) is \(\frac{a+b}{2}\). In an arithmetic sequence, every term (except the first and last) is the arithmetic mean of its two neighbours. This property serves as both a definition test and a practical tool.

\[ a_n = \frac{a_{n-1} + a_{n+1}}{2} \]

Inserting arithmetic means: To insert k arithmetic means between two values \(a\) and \(b\), the common difference is \(d = \frac{b - a}{k + 1}\). For instance, inserting 3 means between 2 and 18: \(d = \frac{18 - 2}{4} = 4\), giving the sequence 2, 6, 10, 14, 18.

Real-World Applications of Arithmetic Sequences

Arithmetic progressions are far more than abstract algebra exercises. They model any situation where a quantity changes by a fixed amount per period.

Finance and Savings

A person who saves $200 in month one and increases the deposit by $50 each month follows the AP 200, 250, 300, 350, ... The sum formula reveals total savings after any number of months. Straight-line depreciation in accounting is another classic AP application: an asset loses the same dollar amount each year.

Construction and Architecture

Staircase design relies on arithmetic sequences: each step rises by the same height (the riser). Fence posts placed at equal intervals, tiered seating in stadiums, and evenly spaced floor tiles are all governed by APs.

Physics and Kinematics

Under constant acceleration, the distances covered in successive equal time intervals form an arithmetic sequence. A freely falling object covers distances proportional to 1, 3, 5, 7, ... in successive seconds, an AP with \(d = 2\) (in units of \(\tfrac{1}{2}g\)).

Computer Science

Loop counters that increment by a fixed step, memory address offsets, and certain hashing strategies use arithmetic progressions. The total number of operations in nested loops often reduces to the sum of an AP, making the sum formula crucial for algorithm analysis.

Music and Nature

Equally tempered tuning adjustments on stringed instruments and evenly spaced calendar events (bi-weekly pay, quarterly reports) are modelled by arithmetic sequences.

Derivation of the Arithmetic Series Sum Formula

Understanding why the sum formula works deepens mathematical maturity. The proof attributed to young Gauss proceeds by writing the series twice, once forward and once backward, then adding term by term.

Write the sum in two orders:

\[ S_n = a_1 + (a_1+d) + (a_1+2d) + \cdots + a_n \] \[ S_n = a_n + (a_n-d) + (a_n-2d) + \cdots + a_1 \]

Adding these two equations column by column, every pair sums to \(a_1 + a_n\), and there are n such pairs:

\[ 2S_n = n(a_1 + a_n) \implies S_n = \frac{n}{2}(a_1 + a_n) \]

Substituting \(a_n = a_1 + (n-1)d\) yields the alternative form \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\). Both expressions are algebraically identical.

Expert Tips for Solving Arithmetic Sequence Problems

1. Verify the sequence is actually arithmetic

Always compute at least two consecutive differences before assuming a constant d. A single difference can be misleading if the sequence is quadratic or geometric.

2. Organise givens before choosing a formula

Write down \(a_1\), \(d\), \(n\), \(a_n\), and \(S_n\). Identify which three you know and which one you need. The right formula usually becomes obvious.

3. Watch for off-by-one errors

The factor \((n-1)\) means the first term is multiplied by zero differences. A common mistake is writing \(a_n = a_1 + nd\) instead of \(a_1 + (n-1)d\).

4. Use the mean property as a shortcut

If three consecutive terms are \(a\), \(b\), \(c\), then \(b = \frac{a+c}{2}\). This can quickly solve for a missing middle term without the full formula.

5. Handle negative differences carefully

Misplacing a negative sign is the most common arithmetic error. When \(d < 0\), double-check every substitution step.

Common Mistakes and How to Avoid Them

MistakeWhy It HappensCorrect Approach
Using \(nd\) instead of \((n-1)d\)Forgetting the first term has 0 differencesAlways write \(a_n = a_1 + (n-1)d\)
Wrong sign on dSubtracting in the wrong orderCalculate \(d = a_2 - a_1\), not \(a_1 - a_2\)
Confusing term value with term position\(a_n\) is the value; \(n\) is the positionLabel each clearly before substituting
Applying AP formulas to a geometric sequenceNot verifying constant differenceCheck at least two differences first
Rounding mid-calculationPremature rounding introduces errorCarry fractions through; round only at the end

Arithmetic vs Geometric Sequences: Complete Comparison

FeatureArithmetic SequenceGeometric Sequence
Defining operationAdd constant difference (d)Multiply by constant ratio (r)
General term\(a_n = a_1 + (n-1)d\)\(a_n = a_1 \cdot r^{n-1}\)
Sum of n terms\(\frac{n}{2}[2a_1 + (n-1)d]\)\(a_1 \frac{1 - r^n}{1 - r}\) (for \(r \ne 1\))
Graph shapeLinear (straight line)Exponential (curve)
Infinite sumDiverges (unless \(d=0\))Converges when \(|r| < 1\)
Mean propertyArithmetic mean of neighboursGeometric mean of neighbours
Example3, 7, 11, 15, 193, 6, 12, 24, 48

Frequently Asked Questions

What is an arithmetic sequence?

An arithmetic sequence is a list of numbers where each successive term is found by adding a fixed constant, called the common difference, to the previous term. For example, 5, 10, 15, 20 is arithmetic with \(d = 5\).

How do you find the nth term of an arithmetic sequence?

Apply \(a_n = a_1 + (n-1)d\). Substitute the first term, the common difference, and the desired position. For the 10th term of 2, 5, 8: \(a_{10} = 2 + 9 \times 3 = 29\).

What is the formula for the sum of an arithmetic series?

Use \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\) or \(S_n = \frac{n}{2}(a_1 + a_n)\). Both give the same result; choose whichever matches the values you already know.

How do you calculate the common difference?

Subtract any term from the next term: \(d = a_{n+1} - a_n\). Alternatively, use \(d = \frac{a_n - a_1}{n - 1}\) if you know two non-consecutive terms and their positions.

Can the common difference be negative or zero?

Yes. A negative d produces a decreasing sequence (e.g., 30, 25, 20, 15 with \(d = -5\)). A zero d gives a constant sequence where every term equals \(a_1\).

What is the difference between an arithmetic sequence and a geometric sequence?

Arithmetic sequences add a constant; geometric sequences multiply by a constant. Their formulas, graph shapes, and convergence behaviours are entirely different, as shown in the comparison table above.

How do I find the number of terms in a finite arithmetic sequence?

Use \(n = \frac{a_n - a_1}{d} + 1\). If the result is not a positive integer, the intended last term does not belong to the sequence.

What is the recursive formula for an arithmetic sequence?

The recursive definition is \(a_n = a_{n-1} + d\) with a given starting value \(a_1\). Each term is computed from the preceding one rather than directly from n.

Official Educational Resources

For deeper exploration of arithmetic sequences, consult these authoritative references:

Master Arithmetic Sequences with Confidence

Arithmetic sequences connect school-level algebra with real-world problem solving. The six core formulas covered above, the nth term, sum of series, common difference, first term, number of terms, and recursive relation, are sufficient to tackle virtually any arithmetic progression question on standardised tests and university courses alike.

Bookmark this page and return whenever you need a quick calculation or a refresher on the underlying theory. Consistent practice with varied problems, especially those involving negative differences, fractional terms, and word-problem contexts, builds lasting fluency.

Related on RevisionTown

Start here (prerequisites)

Practice and application

Next steps

Shares: