The Number That Makes an Arithmetic Sequence "Arithmetic"
Some sequences grow by multiplying. Others grow by adding the same amount each step. The second kind — arithmetic sequences — are completely characterised by a single number: the common difference.
Once you know the first term and the common difference, you know every term of the sequence forever. The common difference is the "rate" of an arithmetic progression — its slope.
What the Common Difference Is
The common difference $d$ of an arithmetic progression (AP) is the constant difference between any term and its predecessor:
$$d = a_n - a_{n-1} = a_{n+1} - a_n = \cdots.$$
This must be the same value for every consecutive pair. If the differences vary, the sequence is not arithmetic.
Examples. $;5, 8, 11, 14, 17, \ldots$ — $d = 3$. $;;100, 95, 90, 85, \ldots$ — $d = -5$. $;;2.5, 4.0, 5.5, 7.0, \ldots$ — $d = 1.5$. $;;7, 7, 7, 7, \ldots$ — $d = 0$ (constant sequence).
Quick facts.
Formula: $d = a_n - a_{n-1}$.
Sign: can be positive (increasing AP), negative (decreasing AP), or zero (constant AP).
Value: any real number — integer, fraction, decimal, irrational.
$n$th term of AP: $a_n = a_1 + (n - 1)d$.
Sum of first $n$ terms: $S_n = \tfrac{n}{2}(2a_1 + (n-1)d) = \tfrac{n}{2}(a_1 + a_n)$.
Two non-adjacent terms: $d = (a_m - a_n) / (m - n)$.
Grade introduced: CBSE Class 10 (arithmetic progressions); CCSS-M HSF-LE.A.2 (construct arithmetic and exponential functions); NCERT Class 10 Chapter 5 — Arithmetic Progressions.
How to Find the Common Difference
Case 1 — Adjacent terms given
Subtract any term from the next.
Example. For $4, 9, 14, 19, \ldots$: $d = 9 - 4 = 5$. (Confirm: $14 - 9 = 5$ ✓.)
Case 2 — Two non-adjacent terms given
Use $d = (a_m - a_n) / (m - n)$.
Example. The 3rd term is 11 and the 7th term is 23. Then $d = (23 - 11)/(7 - 3) = 12/4 = 3$.
Case 3 — From the $n$th-term formula
If $a_n = 5n + 2$, then $d = $ coefficient of $n$ in linear form $= 5$.
The $n$th Term Formula
The general term of an arithmetic progression is
$$a_n = a_1 + (n - 1) d.$$
Derivation. $a_2 = a_1 + d$. $a_3 = a_2 + d = a_1 + 2d$. By induction, $a_n = a_1 + (n - 1)d$.
The exponent on $d$ is $n - 1$ because the first term involves zero additions: $a_1 = a_1 + 0 \cdot d$.
Three Worked Examples — Quick, Standard, Stretch
Quick. Find the common difference of $3, 7, 11, 15, 19, \ldots$.
$$d = 7 - 3 = 4.$$
Confirm: $11 - 7 = 4$, $15 - 11 = 4$. Constant.
Final answer: $d = 4$.
Standard (Wrong Path First — Where Students Trip Up). The 4th term of an AP is 17 and the 9th term is 32. Find the common difference.
The wrong path. The rusher subtracts: $32 - 17 = 15$. They report $d = 15$.
The flaw: the two terms are not adjacent. $32 - 17$ is the total increase over five steps (from term 4 to term 9), not the per-step increase.
The rescue. Apply the non-adjacent-terms formula.
$$d = \frac{a_9 - a_4}{9 - 4} = \frac{32 - 17}{5} = 3.$$
Final answer: $d = 3$.
Stretch. The first term of an AP is 5 and the sum of the first 10 terms is 185. Find the common difference.
Use the sum formula $S_n = \tfrac{n}{2}(2a_1 + (n - 1)d)$ with $n = 10$, $a_1 = 5$.
$$185 = \tfrac{10}{2}(2 \cdot 5 + 9d).$$ $$185 = 5(10 + 9d).$$ $$37 = 10 + 9d.$$ $$9d = 27 \implies d = 3.$$
Final answer: $d = 3$.
Why the Common Difference Matters — Beyond Pattern Recognition
The common difference is the slope of an arithmetic progression. Once you know it, you can compute any term, any sum, any prediction about the sequence.
Salary increments. A starting salary of $40{,}000$ with a yearly increment of $3{,}000$ is an AP with $a_1 = 40000$, $d = 3000$. Year 10 salary is $a_{10} = 40000 + 9 \cdot 3000 = 67{,}000$.
Loan repayments. Fixed-instalment loans (the principal portion of each instalment) form an AP.
Stadium seating. Rows expanding by a constant number of seats each row form an AP. Total seats over $n$ rows is the AP sum.
Counting problems. "How many integers between 100 and 999 are divisible by 7?" — the divisible-by-7 integers form an AP with $d = 7$.
Linear motion. An object moving at constant velocity passes equally-spaced markers at equal time intervals — an AP in position.
The destination, in every direction: any time something grows by the same amount per step, the common difference is the rate and the AP is the model.
The Tripping Points of Common Difference
1. Subtracting non-adjacent terms without averaging.
Where it slips in: Terms at positions 4 and 9 — student subtracts directly.
Don't do this: Treat $a_9 - a_4$ as the common difference.
The correct way: $d = (a_9 - a_4) / (9 - 4)$. Divide the difference by the number of steps between the terms.
2. Confusing the common difference with the common ratio.
Where it slips in: Asked for the common difference of $2, 6, 18, 54, \ldots$, student reports $d = 4$ (the first difference).
Don't do this: Treat any constant pattern as arithmetic.
The correct way: Check whether the difference is constant. $6 - 2 = 4$ but $18 - 6 = 12$ — differences vary, so this is not an AP. The ratio is constant — a GP.
3. Forgetting that $d$ can be zero.
Where it slips in: The sequence $5, 5, 5, 5, \ldots$ — student says "this can't be an AP."
Don't do this: Restrict $d$ to non-zero values.
The correct way: $d = 0$ is allowed. The sequence is a constant AP.
4. Misapplying the index in the $n$th-term formula.
Where it slips in: Student writes $a_n = a_1 + nd$ instead of $a_n = a_1 + (n - 1)d$.
Don't do this: Drop the $-1$ in the index.
The correct way: The first term involves zero additions, so $a_1 = a_1 + 0 \cdot d = a_1 + (1 - 1)d$. The formula has $n - 1$, not $n$.
The real-world version. In 1801, the astronomer Carl Friedrich Gauss — then 24 years old — predicted the orbital position of the dwarf planet Ceres after only 40 days of telescope observations. His prediction relied on detecting the common difference in the apparent angular position from one observation to the next, then extrapolating with the AP formula across the months when Ceres was hidden behind the Sun. Astronomers found Ceres exactly where Gauss said it would be. The arithmetic-progression rate of motion, applied across a hidden interval, is what made the recovery possible.
The Mathematicians Who Worked With Arithmetic Progressions
Aryabhata (476–550 CE, India) gave the formula for the sum of an arithmetic progression in Aryabhatiya (499 CE), centuries before similar formulas appeared in Europe.
Carl Friedrich Gauss (1777–1855, Germany) famously summed $1 + 2 + \cdots + 100 = 5050$ in seconds as a child by recognising the AP sum formula $S_n = n(a_1 + a_n)/2$.
Pierre de Fermat (1607–1665, France) worked with arithmetic progressions in number theory, including the famous Fermat's theorem on arithmetic progressions (every AP with first term $a$ and common difference $d$ coprime to $a$ contains infinitely many primes — eventually proved by Dirichlet in 1837).
Conclusion
The common difference $d$ is the constant added to each term of an arithmetic progression.
For adjacent terms, $d = a_n - a_{n-1}$.
For non-adjacent terms, $d = (a_m - a_n) / (m - n)$.
The $n$th term is $a_n = a_1 + (n - 1) d$.
The single most common mistake is subtracting non-adjacent terms without dividing by the index gap.
Sharpen Your Common Difference — Three Practice Problems
Find the common difference of $-2, 1, 4, 7, 10, \ldots$.
The 5th term of an AP is 22, and the 12th term is 50. Find the common difference.
The first term is 7 and the sum of the first 8 terms is 92. Find the common difference.
Want a live Bhanzu trainer to walk through more arithmetic-progression problems? Book a free demo class — online globally.
Was this article helpful?
Your feedback helps us write better content