Add the same step every time, and the dots fall in a straight line.
Climb a staircase whose steps are all the same height and you never have to think — each move lifts you by exactly the same amount. An arithmetic sequence is that staircase written as numbers: a starting value, then a fixed step d added again and again. Because the step never changes, the whole sequence is decided by just two numbers — the first term a₁ and the common difference d — and the dots you plot land on a perfectly straight line. In this lesson we turn that picture into a formula for any term, learn the gentle balance hidden in three terms in a row, and then perform the trick a ten-year-old Carl Gauss used to add 1 + 2 + ⋯ + 100 in seconds. By the end you will hold two formulas — one for the term aₙ and one for the running total Sₙ — and know exactly which is which.
Look again at 3, 7, 11, 15, 19, …. Subtract each term from the one after it: 7 − 3 = 4, 11 − 7 = 4, 15 − 11 = 4. The gap is the same every time. That repeating gap is the whole idea.
A sequence {aₙ} is arithmetic when consecutive terms differ by the same constant, called the common difference d:
d = an+1 − aₙ (the same number for every n).
If d > 0 the sequence climbs; if d < 0 it falls; if d = 0 every term is the same. The test is one-sided in time but two-sided in meaning: every term is its predecessor plus d, and equally its successor minus d.
To check a sequence, difference it. The list 2, 5, 8, 11, 14 gives the gaps 3, 3, 3, 3 — arithmetic with d = 3. The list 1, 2, 4, 8, 16 gives gaps 1, 2, 4, 8 — not arithmetic (those gaps double; that is a geometric sequence, our next lesson). Differencing is the single test: equal differences in, arithmetic out.
The sequence 23, 18, 13, 8, 3, −2, … has 18 − 23 = −5, and indeed every gap is −5. So a₁ = 23, d = −5: a steady descent of 5 a step. Note the minus sign in d is doing real work — it points the staircase downward.
To reach the n-th term, start at a₁ and take steps of size d. How many steps? Getting to a₂ takes one step, to a₃ takes two, to aₙ takes (n − 1) steps. Multiply the step by how many you take and add to the start:
aₙ = a₁ + (n − 1)d
For 3, 7, 11, 15, … with a₁ = 3, d = 4: aₙ = 3 + 4(n − 1). Then a₁₀ = 3 + 4·9 = 39, and a₁₀₀ = 3 + 4·99 = 399 — no need to list the first ninety-nine terms.
Now read that formula with the eyes of Stage 21. Expand it:
aₙ = d·n + (a₁ − d).
That is exactly the shape of a linear function y = mx + b, with the index n playing the role of x. The slope is the common difference d, and the intercept is a₁ − d. So the points (n, aₙ) are collinear — they all lie on one straight line of slope d. An arithmetic sequence is simply a linear function fed only the whole numbers 1, 2, 3, …; its graph is that line's evenly spaced dots.
The line is a guide: the sequence itself lives only at the whole-number inputs, so plot it as separated dots, never a connected stroke. The straight line tells you they are collinear; it does not mean a₂.₅ exists. There is no "term number 2.5."
Set the first term and the common difference; the dots are the sequence, the faint blue line is aₙ = a₁ + (n−1)d that they sit on. The readout gives the formula, a₁₀, and S₁₀.
Take any three terms in a row, say 3, 7, 11. The middle one, 7, sits exactly halfway between its neighbours: 7 − 3 = 4 and 11 − 7 = 4 are equal — that is what "constant difference" means locally. Halfway between two numbers is their average, so the middle term is the average of the outer two.
A number b is the arithmetic mean of a and c exactly when
2b = a + c, equivalently b = a + c2.
In an arithmetic sequence, every middle term is the mean of its two neighbours: 2aₙ = an−1 + an+1. So to insert a term between a and c so all three are arithmetic, use their average.
Why is this the right notion of "middle"? Because b − a = c − b rearranges into 2b = a + c. Equal gaps on both sides is the mean condition — they say the same thing. Insert b between 4 and 16 arithmetically and you must take b = (4 + 16)/2 = 10, giving the run 4, 10, 16 with common difference 6.
If 3, x, 17 are three consecutive terms of an arithmetic sequence, then x is their mean: x = (3 + 17)/2 = 10. Check: 10 − 3 = 7 and 17 − 10 = 7 — equal. The common difference of this little run is 7.
The mean was a balance among three terms in a row. The same balance reaches across the whole sequence. Here is the rule, and it is the engine behind the sum formula in the next section.
If the index sums match, the term sums match:
m + n = p + q ⟹ aₘ + aₙ = ap + aq.
In particular, terms equidistant from the ends share one sum: a₁ + aₙ = a₂ + an−1 = a₃ + an−2 = ⋯
The proof is one line of algebra. Each term is a₁ + (index − 1)d, so
aₘ + aₙ = 2a₁ + (m + n − 2)d, ap + aq = 2a₁ + (p + q − 2)d.
If m + n = p + q the right-hand sides are identical, so the two pairs add to the same thing. Intuitively: walking +d on one index and −d on the other keeps the total fixed — what one term gains, its partner loses.
In an arithmetic sequence, a₂ = 5 and a₆ = 17. Since 2 + 6 = 4 + 4, the property gives a₂ + a₆ = a₄ + a₄ = 2a₄. So 2a₄ = 5 + 17 = 22, hence a₄ = 11 — the mean of a₂ and a₆, no formula for d needed.
Now the famous trick. We want Sₙ = a₁ + a₂ + ⋯ + aₙ, the running total of the first n terms. Story has it that a schoolteacher told the young Carl Gauss to add the whole numbers from 1 to 100, expecting a long quiet hour. Gauss wrote one number. Here is what he saw.
Write the sum forwards, then write it again backwards underneath, and add the two rows column by column:
Sₙ = a₁ + a₂ + ⋯ + an−1 + aₙ
Sₙ = aₙ + an−1 + ⋯ + a₂ + a₁
2Sₙ = (a₁+aₙ) + (a₂+an−1) + ⋯ + (aₙ+a₁)
By the balance property of the last section, every one of those n columns adds to the same value a₁ + aₙ. So 2Sₙ = n(a₁ + aₙ), and dividing by 2:
Sₙ = n(a₁ + aₙ)2 = na₁ + n(n − 1)d2
The first form — n times the average of the ends — is the one to remember. The second comes from substituting aₙ = a₁ + (n−1)d, handy when you know a₁ and d but not aₙ yet.
Gauss's sum is the special case a₁ = 1, d = 1, n = 100, so aₙ = 100:
1 + 2 + ⋯ + 100 = 100·(1 + 100)2 = 100·1012 = 5050.
For our staircase 3, 7, 11, 15, 19: n = 5, a₁ = 3, a₅ = 19, so S₅ = 5·(3 + 19)/2 = 5·11 = 55. And the first ten terms (a₁₀ = 39) sum to S₁₀ = 10·(3 + 39)/2 = 210. Drive the rectangle below and watch two copies of Sₙ tile a perfect block.
Dial how many terms n you sum (the sequence is fixed at a₁ = 3, d = 4). The amber bars are the terms; the blue bars are the reversed copy stacked on top. Every column locks to the same height a₁ + aₙ — read off 2Sₙ = n(a₁+aₙ).
Keep aₙ and Sₙ apart. The term a₁₀ = 39 is one value; the running total S₁₀ = 210 sums all ten. "The 10th term" and "the sum of the first 10 terms" are different questions — confusing them is the most common slip in the whole topic. And mind that d is the step, not the first term: in 3, 7, 11, … the step is 4, while a₁ is 3.
One more reading of the sum, this time through Stage 23. Take Sₙ = na₁ + n(n−1)d/2 and expand it as a function of n:
Sₙ = d2n2 + (a₁ − d2)n.
That is a quadratic in n — a parabola, with leading coefficient d/2 and no constant term (S₀ = 0). The term aₙ was linear in n; the sum Sₙ is quadratic. That is the cleanest way to remember which is which.
The parabola earns its keep when d < 0. Then the staircase descends: early terms are positive, later ones turn negative, and the leading coefficient d/2 is negative — so the parabola opens downward and Sₙ has a maximum at its vertex. The running total grows while the terms are still positive, then peaks the moment they cross zero, then shrinks.
Take a₁ = 20, d = −3: the terms 20, 17, 14, 11, 8, 5, 2, −1, … stay positive through n = 7 (a₇ = 2) and go negative at n = 8 (a₈ = −1). So the sum keeps climbing until you have added the last positive term, then falls. The largest total is S₇ = 7·(20 + 2)/2 = 77. The vertex of the parabola Sₙ sits right where the terms change sign — find the last positive term and stop.
Fix a₁ = 20 and dial the common difference d. The green dots are the partial sums Sₙ; they sit exactly on the blue parabola Sₙ(n). When d < 0 the parabola opens down and the sum has a maximum — read where the dots turn around.
An arithmetic sequence is decided by two numbers, a₁ and d. From them, one formula gives any term and another gives any running total — and they live at different "degrees" in n.
| Object | Formula | Shape in n | Reading |
|---|---|---|---|
| common difference | d = an+1 − aₙ | constant | the fixed step |
| the term aₙ | aₙ = a₁ + (n−1)d | linear (a line) | slope d, dots collinear |
| arithmetic mean | 2b = a + c | — | middle = average of neighbours |
| balance | m+n=p+q ⟹ aₘ+aₙ=ap+aq | — | ends pair to one sum |
| the sum Sₙ | Sₙ = n(a₁+aₙ)/2 | quadratic (a parabola) | n × average of the ends |
| trap to avoid | aₙ ≠ Sₙ | — | term vs running total |
Next we keep the same story but swap the operation: instead of adding the same step, multiply by the same ratio. The dots leave the straight line and curve like an exponential — that is the geometric sequence, and its running total creeps toward a limit.
a₁ = 5 and d = 9 − 5 = 4. So aₙ = 5 + 4(n − 1) = 4n + 1. Then a₂₀ = 4·20 + 1 = 81 (or 5 + 4·19 = 81).
From a₃ to a₇ is 4 steps, and the value rose 24 − 12 = 12, so d = 12/4 = 3. Back up from a₃: a₁ = a₃ − 2d = 12 − 6 = 6. (Check: 6, 9, 12, 15, 18, 21, 24 — a₇ = 24. ✓)
The middle term is the arithmetic mean: (7 + 23)/2 = 15. The run is 7, 15, 23 with common difference 23 − 15 = 8.
Solve aₙ = 59: 2 + 3(n − 1) = 59 ⟹ 3(n − 1) = 57 ⟹ n = 20 terms. Then S₂₀ = 20·(2 + 59)/2 = 20·61/2 = 10·61 = 610.
Here a₁ = 2, d = 2, aₙ = 100, so 2 + 2(n − 1) = 100 ⟹ n = 50. Sum = 50·(2 + 100)/2 = 50·51 = 2550. (Sanity check: it is twice 1 + 2 + ⋯ + 50 = 2·1275 = 2550. ✓)
aₙ = 25 − 4(n − 1) = 29 − 4n. It is positive while 29 − 4n > 0, i.e. n < 7.25, so the last positive term is a₇ = 29 − 28 = 1 (a₈ = −3 is already negative). The sum is largest at n = 7: S₇ = 7·(25 + 1)/2 = 7·13 = 91. Adding a₈ would only shrink the total.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops arithmetic sequences as linear functions of the position n and derives the sum formula by the reverse-and-add argument. It targets CCSS HSF-BF.A.2 (write arithmetic sequences both recursively, an+1 = aₙ + d, and with an explicit formula, aₙ = a₁ + (n−1)d, and translate between the two) and HSF-LE.A.2 (construct arithmetic sequences given a description, a graph of collinear points, or two terms). The closed form for Sₙ = n(a₁+aₙ)/2 supports HSA-SSE.B.4's reasoning about finite series, and recognizing Sₙ as a quadratic in n connects back to HSF-IF.C.7a (the parabola and its maximum). Throughout, students practice MP7 (look for structure) when they pair terms that balance around the middle, and MP8 (regularity in repeated reasoning) when they see the constant difference become a slope.