The arithmetic sequence formula refers to two related formulas: one for finding any specific term in the sequence (the nth term), and one for finding the sum of the first n terms. The nth term formula is aₙ = a₁ + (n−1)d, and the sum formula is Sₙ = n/2 [2a₁ + (n−1)d]. An arithmetic sequence is a list of numbers where each term differs from the previous one by a fixed amount called the common difference, written as d.
The Two Arithmetic Sequence Formulas
Formula for the nth Term
aₙ = a₁ + (n−1)d
Use this formula when you need a specific term in the sequence without listing every term before it.
Formula for the Sum of n Terms
Two equivalent forms exist. Use the first when you know the common difference; use the second when you already know the last term.
Sₙ = (n/2) × [2a₁ + (n−1)d]
Sₙ = (n/2) × (a₁ + aₙ)
Both produce the same answer. The choice depends on which values are given.
Variable Key
Variable | Meaning |
|---|---|
aₙ | The nth term — the term you're trying to find |
a₁ | The first term of the sequence |
n | The position of the term in the sequence |
d | The common difference between consecutive terms |
Sₙ | The sum of the first n terms |
aₖ | An intermediate term used in alternative formulas |
Why the Formula Uses (n−1), Not n
The (n−1) is the most common source of confusion in this formula. The reason it appears, rather than n, becomes clear when the first few terms are written in expanded form.
To get from the 1st term to the 2nd, you add d once. To reach the 3rd, you add d twice. To reach the 5th, you add d four times — not five. The (n−1) counts the jumps between terms, not the terms themselves.
The pattern shows clearly when terms are listed:
a₁ = a₁
a₂ = a₁ + d
a₃ = a₁ + 2d
a₄ = a₁ + 3d
a₁₀ = a₁ + 9d
a₁₀₀ = a₁ + 99d
The coefficient on d is always one less than the position number. That is the (n−1).
How the Sum Formula Is Derived
The sum formula comes from a reverse-and-add trick attributed to Carl Friedrich Gauss as a young student. Asked by his teacher to add the integers from 1 to 100, he found a shortcut in seconds.
The shortcut: pair the first number with the last, the second with the second-to-last, and so on.
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
... and so on, down to 50 + 51 = 101.
Every pair sums to 101. There are 50 such pairs. Total: 50 × 101 = 5,050.
To turn this trick into a formula, write the sum twice — once forward, once backward — and add term by term.
Forward:
S = a₁ + (a₁ + d) + (a₁ + 2d) + ... + aₙ
Backward:
S = aₙ + (aₙ − d) + (aₙ − 2d) + ... + a₁
Adding the two equations column by column, every column sums to (a₁ + aₙ). There are n columns, so:
2S = n × (a₁ + aₙ)
Sₙ = (n/2) × (a₁ + aₙ)
Substituting aₙ = a₁ + (n−1)d into this equation gives the second form:
Sₙ = (n/2) × [2a₁ + (n−1)d]
The two sum formulas are not separate rules. They are the same formula expressed in terms of different known values.
Other Useful Formulas in the Same Family
The two main formulas (nth term and sum) are part of a small family. Each formula in the family rearranges the same underlying relationship to solve for a different unknown.
Formula | What It Finds | When to Use It |
|---|---|---|
aₙ = a₁ + (n−1)d | The nth term | a₁ and d are known |
aₙ = aₖ + (n−k)d | The nth term from any other known term | Two non-consecutive terms are known |
d = (aₙ − a₁) / (n−1) | The common difference | First and any other term are known |
n = (aₙ − a₁) / d + 1 | The number of terms | First, last, and d are known |
Sₙ = (n/2) × [2a₁ + (n−1)d] | The sum of n terms | a₁ and d are known |
Sₙ = (n/2) × (a₁ + aₙ) | The sum of n terms | a₁ and aₙ are known |
Worked Examples
Example 1 — Find the nth Term
Find the 20th term of the sequence 4, 11, 18, 25, ...
a₁ = 4, d = 7, n = 20.
a₂₀ = 4 + (20 − 1)(7) = 4 + 133 = 137
Example 2 — Find the Sum of n Terms
Find the sum of the first 25 terms of 3, 7, 11, 15, ...
a₁ = 3, d = 4, n = 25.
S₂₅ = (25/2) × [2(3) + (25 − 1)(4)] = (25/2) × [6 + 96] = (25/2) × 102 = 1,275
Example 3 — Negative Common Difference
Find the 12th term of the sequence 50, 44, 38, 32, ...
a₁ = 50, d = −6, n = 12.
a₁₂ = 50 + (12 − 1)(−6) = 50 − 66 = −16
The common difference is negative because the sequence decreases.
Example 4 — Fractional Common Difference
Find the 30th term of 1.5, 2.0, 2.5, 3.0, ...
a₁ = 1.5, d = 0.5, n = 30.
a₃₀ = 1.5 + (30 − 1)(0.5) = 1.5 + 14.5 = 16
Example 5 — Two Non-Consecutive Terms
The 5th term of an arithmetic sequence is 22. The 12th term is 50. Find d and a₁.
Use aₙ = aₖ + (n − k)d with the two known terms:
50 = 22 + (12 − 5)d 28 = 7d d = 4
Then find a₁ using the nth term formula:
22 = a₁ + (5 − 1)(4) 22 = a₁ + 16 a₁ = 6
Example 6 — Find the Number of Terms
The first term is 7, the last term is 187, and the common difference is 5. How many terms are in the sequence?
n = (187 − 7) / 5 + 1 = 36 + 1 = 37
Sigma Notation: The Sum Formula in One Line
The sum formula can be written compactly using sigma notation, the standard notation for sums in higher-level mathematics.
Sₙ = ∑ (k=1 to n) [a₁ + (k−1)d]
The Greek letter Σ (sigma) means "add up everything that follows." The "k = 1" below sigma marks where the index starts. The "n" above marks where it stops. The expression to the right is what gets evaluated at each value of k and added to the running total.
Mini-example. Evaluate ∑ (k=1 to 5) (2k + 1):
k | 2k + 1 |
|---|---|
1 | 3 |
2 | 5 |
3 | 7 |
4 | 9 |
5 | 11 |
Sum = 3 + 5 + 7 + 9 + 11 = 35
Sigma notation appears from Class 11 (NCERT), high school Algebra 2 (CCSS), and AS-Level Maths (UK) onwards.
Common Mistakes
Three errors account for most lost marks on arithmetic sequence problems.
The first is using n in place of (n−1). A student finds the 10th term by adding 10 copies of d to a₁, when it should be 9. Remember that d is the size of each jump, and there are always one fewer jumps than terms.
The second is the wrong sign on d when the sequence decreases. A sequence like 50, 44, 38 has d = −6, not 6. Forgetting the negative sign sends every subsequent term in the wrong direction and produces an answer with the wrong magnitude.
The third is confusing aₙ with Sₙ. A question asks for the 10th term and the student calculates the sum of the first 10 terms instead. Both formulas use the same inputs (a₁, d, n), so the slip is easy to make. Check what the question is asking before substituting.
Where You'll See This Formula in Your Curriculum
The arithmetic sequence formula appears across major school curricula, almost always in the high school years.
Curriculum | Where It Appears |
|---|---|
NCERT (India) | Class 10, Chapter 5: Arithmetic Progressions |
CCSS (United States) | High School — HSF.BF.A.2, HSF.LE.A.2 |
GCSE (United Kingdom) | Foundation and Higher tier — Sequences |
IB MYP / DP | MYP Year 4–5; DP Mathematics: Analysis & Approaches SL |
Cambridge IGCSE | Extended Maths — Sequences |
Where the Formula Shows Up in Real Life
Arithmetic sequences describe any process where something changes by a fixed amount at regular intervals.
Salary increments. A starting salary with a fixed annual raise forms an arithmetic sequence. Total earnings over n years equal Sₙ.
Loan repayment with fixed principal. When the principal portion paid each month stays constant, the monthly interest charged forms an arithmetic sequence (decreasing).
Stadium and theatre seating. Rows of seats where each row adds the same number of additional seats. Total seating capacity equals Sₙ.
Constant deceleration. A vehicle braking at a constant rate covers distances each second that form a decreasing arithmetic sequence.
Stacking. Cans stacked in a triangular display, with each row containing one more can than the row above, follow an arithmetic sequence with d = 1.
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