Arithmetic Sequence Calculator
Free Arithmetic sequence Calculator for sequences. Enter values to get step-by-step solutions with formulas and graphs.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator
Arithmetic Sequence Calculator
Calculator
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Formula: an = a1 + (n-1)d | Sn = (n/2)(2a1 + (n-1)d)
Worked example — 20th row: 72 seats | Total: 870 seats | Mean: 43.5 per row
Formula
an = a1 + (n-1)d | Sn = (n/2)(2a1 + (n-1)d)
Where an is the nth term, a1 is the first term, d is the common difference, n is the number of terms, and Sn is the sum of the first n terms. The nth term formula finds any specific term, while the sum formula calculates the total of all terms from the first through the nth.
Worked Examples
Example 1: Stadium Seating Layout
Problem:A stadium has 20 rows. The first row has 15 seats and each subsequent row has 3 more seats. Find the number of seats in the 20th row and the total seats.
Solution:a1 = 15, d = 3, n = 20 20th row: a20 = 15 + (20-1) x 3 = 15 + 57 = 72 seats Total seats: S20 = (20/2) x (15 + 72) = 10 x 87 = 870 Arithmetic mean = 870 / 20 = 43.5 seats per row
Result:20th row: 72 seats | Total: 870 seats | Mean: 43.5 per row
Example 2: Salary with Annual Raises
Problem:An employee starts at $45,000 and receives a $2,500 raise each year. What is the salary in year 10 and total earnings over 10 years?
Solution:a1 = 45000, d = 2500, n = 10 Year 10 salary: a10 = 45000 + (10-1) x 2500 = 45000 + 22500 = $67,500 Total earnings: S10 = (10/2) x (45000 + 67500) = 5 x 112500 = $562,500 Average salary = $562,500 / 10 = $56,250
Result:Year 10 salary: $67,500 | Total 10-year earnings: $562,500
Frequently Asked Questions
What is an arithmetic sequence and what defines it?
An arithmetic sequence (also called an arithmetic progression) is an ordered list of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by d. For example, 2, 5, 8, 11, 14 is an arithmetic sequence with first term a1 = 2 and common difference d = 3. Each term equals the previous term plus d. Arithmetic sequences appear naturally in everyday scenarios like counting by fives, monthly salary increments, or evenly spaced fence posts. They are one of the most fundamental concepts in mathematics and serve as building blocks for more advanced topics.
What is the formula for the nth term of an arithmetic sequence?
The nth term formula is: an = a1 + (n - 1) x d, where a1 is the first term, d is the common difference, and n is the position number. This formula lets you find any term without listing all preceding terms. For example, in the sequence 7, 12, 17, 22, ... the 100th term is 7 + (100 - 1) x 5 = 7 + 495 = 502. The formula can also work backwards: if you know a term value and want its position, rearrange to n = ((an - a1) / d) + 1. This is useful for determining whether a specific number belongs to a given arithmetic sequence.
How do I calculate the sum of an arithmetic sequence?
The sum of the first n terms uses the formula: Sn = (n / 2) x (2a1 + (n - 1) x d), or equivalently Sn = (n / 2) x (a1 + an) where an is the last term. This formula was famously discovered by young Carl Friedrich Gauss when asked to sum the numbers 1 through 100. He recognized that pairing the first and last terms (1 + 100 = 101, 2 + 99 = 101, etc.) creates 50 pairs of 101, giving 5,050. The formula generalizes this pairing technique to any arithmetic sequence. It is one of the most elegant and practical formulas in elementary mathematics.
What is the common difference and how do I find it?
The common difference d is the constant value added to each term to get the next term. Calculate it by subtracting any term from the term that follows it: d = a(n+1) - an. For the sequence 14, 9, 4, -1, -6, the common difference is 9 - 14 = -5. A positive d creates an increasing sequence, a negative d creates a decreasing sequence, and d = 0 produces a constant sequence. If you are given non-consecutive terms, you can find d by dividing the difference between terms by the number of gaps. For instance, if a3 = 11 and a7 = 27, then d = (27 - 11) / (7 - 3) = 16 / 4 = 4.
How are arithmetic sequences used in real-life applications?
Arithmetic sequences model situations with constant rates of change. Linear depreciation uses them when an asset loses the same dollar amount each year: a $50,000 machine depreciating $5,000 annually follows the sequence 50000, 45000, 40000, and so on. Salary schedules with fixed annual raises form arithmetic sequences. Stacking objects in rows where each row has one more item than the previous creates arithmetic sequences. Seating arrangements in theaters (rows getting wider by a fixed number of seats), mortgage amortization with fixed principal payments, and drug dosage accumulation at regular intervals all involve arithmetic progressions.
What is the arithmetic mean and how does it relate to the sequence?
The arithmetic mean of an arithmetic sequence equals the average of the first and last terms: Mean = (a1 + an) / 2. This also equals the middle term if the number of terms is odd. For the sequence 5, 8, 11, 14, 17, the arithmetic mean is (5 + 17) / 2 = 11, which is indeed the middle (third) term. The arithmetic mean has an important property: any term in an arithmetic sequence is the arithmetic mean of its two neighbors. That is, an = (a(n-1) + a(n+1)) / 2. This property provides a quick way to verify that a sequence is arithmetic and to find missing terms within the sequence.
How do I find missing terms in an arithmetic sequence?
To find missing terms, first determine the common difference from any two known consecutive terms or from any two terms and their positions. If you know a3 = 10 and a7 = 26, then d = (26 - 10) / (7 - 3) = 4. Then use a1 = a3 - 2d = 10 - 8 = 2 to find the first term. Now fill in all missing terms: a4 = 14, a5 = 18, a6 = 22. If you have three terms and need to determine which value makes them arithmetic, use the property that the middle term equals the average of the outer two terms. This technique is commonly tested in standardized math examinations and competition problems.
What is the difference between arithmetic and geometric sequences?
Arithmetic sequences have a constant difference between consecutive terms (addition/subtraction pattern), while geometric sequences have a constant ratio between consecutive terms (multiplication/division pattern). Arithmetic: 3, 7, 11, 15 (d = 4, each term adds 4). Geometric: 3, 6, 12, 24 (r = 2, each term multiplies by 2). Arithmetic sequences grow linearly and their graphs are straight lines. Geometric sequences grow exponentially and their graphs are curves. The sum of an infinite arithmetic series always diverges (goes to infinity), while the sum of an infinite geometric series converges when the absolute ratio is less than 1.
Can an arithmetic sequence have negative or fractional common differences?
Absolutely. A negative common difference creates a decreasing sequence. Starting at 100 with d = -7 gives 100, 93, 86, 79, 72, and so on. Fractional common differences are equally valid: starting at 1 with d = 0.5 gives 1, 1.5, 2, 2.5, 3, and so on. You can even have irrational common differences like d = sqrt(2), giving the sequence 0, 1.414, 2.828, 4.243, etc. The nth term and sum formulas work identically regardless of whether d is positive, negative, integer, fractional, or irrational. The only restriction is that d must be a fixed constant for the sequence to be arithmetic.
How do I determine if a given set of numbers forms an arithmetic sequence?
Check whether the difference between consecutive terms is constant throughout the entire set. Calculate d = a2 - a1, then verify that a3 - a2 = d, a4 - a3 = d, and so on for all consecutive pairs. If any pair yields a different difference, the numbers do not form an arithmetic sequence. For large datasets, this can be automated by computing all consecutive differences and checking if they all equal each other within acceptable rounding tolerance. Another quick check uses the property that for any three consecutive terms, the middle term must equal the average of the other two. This approach requires fewer calculations for spot-checking.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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