Arithmetic Sequences and Series

Suppose you drop £30 into a jar today, then £30 more every month without fail. Or you stack a wall of cans: 12 on the bottom row, 11 above, 10 above that. Or you climb stadium seating where each tier holds four more seats than the one below. Every one of these marches up (or down) by the same fixed step each time — and that regular, predictable march is an arithmetic sequence.

The beauty is that you never have to grind through every term. Two tidy formulas do the heavy lifting: one snaps out any term you ask for (the 100th, say) without listing the first 99, and one adds up a whole run of them in a single line. Let's build both.

An arithmetic sequence climbs (or falls) by the same amount at every step. That fixed gap is the common difference d. Starting from a first term a, you keep adding d:

a,\; a + d,\; a + 2d,\; a + 3d,\; \dots

Look at the pattern of little multipliers: term 1 has 0 steps of d, term 2 has 1, term 3 has 2. The term number is always one ahead of the number of steps. This is exactly the linear nth-term rule seen from the front: the nth term is the first term plus (n-1) jumps of d.

a_n = a + (n - 1)d

a is the first term;
d is the common difference (add it to get the next term);
n is the position you want, and (n-1) counts the steps taken to reach it.

Worked example — the 20th term

A sequence starts 3, 7, 11, 15, \dots Find the 20th term.

First read off the ingredients: the first term is a = 3, and each step adds 4, so d = 4. We want n = 20. Drop them into the formula — and mind that it's (n-1), not n:

a_{20} = 3 + (20 - 1)\times 4 = 3 + 19\times 4 = 3 + 76 = 79

The 20th term is 79. Notice we took 19 steps of 4, not 20 — the very first term is where we begin, before any step is taken.

A series is what you get when you add up the terms of a sequence. The sum of the first n terms of an arithmetic sequence is written S_n. We could just substitute and add one term at a time — but there is a far quicker way, and it comes from a lovely trick.

See it built

Here is the trick the young Gauss is said to have used to add 1 + 2 + \dots + 100 in seconds. Write the sum forwards, write it again backwards underneath, and add the two rows: every column makes the same total, (\text{first} + \text{last}). Step through it.

There are n columns, each adding to (\text{first} + \text{last}), and the two rows together count the sum twice. So:

2S_n = n\,(\text{first} + \text{last}) \quad\Longrightarrow\quad S_n = \tfrac{n}{2}\,(\text{first} + \text{last})

Since the last term is a + (n-1)d, substituting gives the form you can use straight from a, d and n:

S_n = \frac{n}{2}\bigl(2a + (n-1)d\bigr)

For 1 + 2 + \dots + 100 that is \tfrac{100}{2}(1 + 100) = 50 \times 101 = 5050. Two formulas, same sum — reach for \tfrac{n}{2}(\text{first}+\text{last}) when you already know both ends, and \tfrac{n}{2}(2a+(n-1)d) when you only know a and d.

Worked example — the sum of the first 50 terms

Take the same sequence 3, 7, 11, 15, \dots (a = 3, d = 4). Add the first 50 terms.

We know a and d but not the last term, so use the second form:

S_{50} = \frac{50}{2}\bigl(2\times 3 + (50 - 1)\times 4\bigr) = 25\,(6 + 196) = 25 \times 202 = 5050

A tidy 5050 again — pure coincidence with Gauss's schoolroom sum, but a nice one. If you happened to spot that the 50th term is 3 + 49\times 4 = 199, the first form agrees: \tfrac{50}{2}(3 + 199) = 25\times 202 = 5050.

Worked example — working backwards

Sometimes you know the total and must find how many terms it took. How many terms of 5, 8, 11, 14, \dots add up to 440?

Here a = 5, d = 3, and S_n = 440; the unknown is n. Put the numbers into the sum formula and tidy up:

440 = \frac{n}{2}\bigl(2\times 5 + (n-1)\times 3\bigr) = \frac{n}{2}\,(3n + 7)

Multiply out into a quadratic:

880 = 3n^2 + 7n \quad\Longrightarrow\quad 3n^2 + 7n - 880 = 0

This factorises as (3n + 55)(n - 16) = 0. The negative root is nonsense for a count of terms, so n = 16: the first 16 terms sum to 440.

Three traps snare almost everyone here:

It's (n-1)d, not nd. The first term already is a — zero steps have been taken. So the 5th term has only four steps of d added, not five. This off-by-one is the single most common mistake. Count the steps, not the terms.
Term or sum? "Find the 20th term" wants a single value (a_n = a + (n-1)d). "Find the sum of the first 20 terms" wants the series (S_n = \tfrac{n}{2}(2a+(n-1)d)). Same sequence, very different answers — read the question twice.
Sign of d. A falling sequence like 20, 17, 14, \dots has d = -3, not 3. Carry the minus sign all the way through.

The story goes that a schoolteacher, wanting a quiet afternoon, told the class to add up every whole number from 1 to 100. Moments later a small boy — Carl Friedrich Gauss — walked up with a single number on his slate: 5050.

He hadn't added them one by one. He'd paired the ends: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, … Fifty pairs, each worth 101, giving 50 \times 101 = 5050. That is exactly the reasoning baked into S_n = \tfrac{n}{2}(\text{first} + \text{last}) — the \tfrac{n}{2} counts the pairs, the bracket is the value of each pair. The same idea totals a stadium's seats, the cans in a stacked display, or the shrinking payments of a loan.

See it explained

Sal Khan derives the arithmetic series formula by averaging the first and last terms.