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Sum of an Arithmetic Sequence

Adding a long run of terms one at a time is slow and easy to slip up on. A neat pairing trick turns the whole job into a single formula you can trust.

MYP 4Standard MathsCoordinate Geometry & SequencesCriteria A · C · D~10 min read

Written by Fayek · MYP student

Why we need a formula

Sometimes you want the total of the first several terms of an arithmetic sequence, not just one term. That total is written \(S_n\), the sum of the first \(n\) terms.

\(S_n\)
The sum of the first \(n\) terms of a sequence: \(S_n = a_1 + a_2 + \cdots + a_n\). For an arithmetic sequence there is a shortcut formula so you never have to add them all by hand.

Adding \(3 + 7 + 11 + \cdots\) for ten terms by hand is doable but slow. For a hundred terms it is a nightmare, so we build a formula instead.

The pairing trick

Take the sum \(1 + 2 + 3 + \cdots + 100\). Pair the first with the last, the second with the second-last, and so on: \(1 + 100 = 101\), \(2 + 99 = 101\), and every pair makes \(101\). There are \(50\) such pairs, so the total is \(50 \times 101 = 5050\).

The key idea is that in an arithmetic sequence, the first term plus the last term equals the second plus the second-last, and so on. Every pair has the same total, \(a + \ell\), where \(\ell\) is the last term. With \(n\) terms you get \(\tfrac{n}{2}\) pairs.

The sum formula

Writing the last term as \(\ell = a + (n-1)d\) and multiplying the number of pairs by the pair total gives the standard formula.

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

Here \(a\) is the first term, \(d\) the common difference and \(n\) how many terms you are adding. If you happen to know the last term \(\ell\) already, the equivalent form \(S_n = \tfrac{n}{2}(a + \ell)\) is quicker.

Using the formula

Pick out \(a\), \(d\) and \(n\), substitute carefully, and simplify inside the bracket before multiplying.

Worked example

Find the sum of the first 10 terms of \(3,\ 7,\ 11,\ 15,\ \ldots\)

1
Read off the values: \(a = 3\), \(d = 7 - 3 = 4\), \(n = 10\).
2
Substitute into \(S_n = \dfrac{n}{2}\bigl(2a + (n-1)d\bigr)\): \(S_{10} = \dfrac{10}{2}\bigl(2(3) + 9(4)\bigr)\).
3
Simplify the bracket: \(2(3) + 9(4) = 6 + 36 = 42\).
4
Multiply: \(S_{10} = 5 \times 42 = 210\).
The sum of the first 10 terms is \(210\).

Do not confuse the sum with the nth term

The nth term formula \(dn + (a-d)\) gives just one term; the sum formula \(\tfrac{n}{2}(2a+(n-1)d)\) adds up the first \(n\) of them. Read the question carefully: "the 10th term" and "the sum of the first 10 terms" are very different numbers.

Check yourself

1. Find the sum of the first 8 terms of \(2,\ 5,\ 8,\ 11,\ \ldots\) +

Here \(a = 2\), \(d = 3\), \(n = 8\). So \(S_8 = \dfrac{8}{2}\bigl(2(2) + 7(3)\bigr) = 4\,(4 + 21) = 4 \times 25 = \mathbf{100}\).

2. Find the sum of the first 20 terms of \(1,\ 2,\ 3,\ 4,\ \ldots\) +

Here \(a = 1\), \(d = 1\), \(n = 20\). So \(S_{20} = \dfrac{20}{2}\bigl(2(1) + 19(1)\bigr) = 10\,(2 + 19) = 10 \times 21 = \mathbf{210}\).

3. Find the sum of the first 12 terms of \(5,\ 9,\ 13,\ 17,\ \ldots\) +

Here \(a = 5\), \(d = 4\), \(n = 12\). So \(S_{12} = \dfrac{12}{2}\bigl(2(5) + 11(4)\bigr) = 6\,(10 + 44) = 6 \times 54 = \mathbf{324}\).

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