Arithmetic Series Formula and How to Find the Sum

An arithmetic series is the sum of the terms of an arithmetic sequence, a list of numbers where each term increases (or decreases) by the same fixed amount, called the common difference \(d\). If the sequence is \(3, 7, 11, 15, 19\), the corresponding arithmetic series is the sum \(3 + 7 + 11 + 15 + 19\).

It is worth keeping the two words straight: a sequence is the ordered list of terms, while a series is what you get when you add those terms together. Once you know a sequence is arithmetic, you can find its sum without adding every term by hand.

Let \(a\) be the first term, \(d\) the common difference, and \(n\) the number of terms. The \(n\)th term of the sequence is \(a_n = a + (n-1)d\). The sum of the first \(n\) terms, written \(S_n\), is given by either of two equivalent forms:

\( S_n = \dfrac{n}{2}\left(a_1 + a_n\right) \)

\( S_n = \dfrac{n}{2}\left(2a + (n-1)d\right) \)

Use the first version when you already know the first and last terms. Use the second version when you know the first term and the common difference but not the last term.

The formula is easy to trust once you see the trick behind it, often credited to a young Gauss adding the numbers \(1\) through \(100\). Write the series forwards and then backwards, term by term:

\( S_n = a_1 + a_2 + a_3 + \dots + a_n \)

\( S_n = a_n + a_{n-1} + a_{n-2} + \dots + a_1 \)

Adding these two lines vertically, each pair of matching terms sums to the same value, \(a_1 + a_n\), and there are \(n\) such pairs. That gives \(2S_n = n(a_1 + a_n)\), and dividing by 2 produces the formula above. This pairing idea is also the reason sigma notation is such a convenient way to write and manipulate these sums.

Because each term of an arithmetic sequence increases by the same amount \(d\), the terms sit on a straight line when plotted against their position number \(n\). This linear pattern is exactly what makes the pairing trick work, the terms are evenly spaced, so the first and last, second and second-to-last, and so on, always average out to the same middle value.

Find the sum of the arithmetic series \(4 + 9 + 14 + \dots + 74\).

First find \(n\). Here \(a_1 = 4\), \(d = 5\), and the last term is \(74\). Using \(a_n = a_1 + (n-1)d\):

\( 74 = 4 + (n-1)(5) \implies 70 = 5(n-1) \implies n = 15 \)

Now apply the sum formula:

\( S_{15} = \dfrac{15}{2}(4 + 74) = \dfrac{15}{2}(78) = 585 \)

Find the sum of the first 20 terms of the arithmetic sequence with first term \(a = 6\) and common difference \(d = -3\).

\( S_{20} = \dfrac{20}{2}\left(2(6) + (20-1)(-3)\right) = 10\left(12 - 57\right) = 10(-45) = -450 \)

Notice the sum is negative because a large negative common difference eventually drags the later terms well below zero.

The most frequent error is mixing up \(n\), the number of terms, with the value of the last term itself. Always solve for \(n\) using \(a_n = a_1 + (n-1)d\) before plugging numbers into the sum formula. Another common slip is forgetting the \(n-1\), not \(n\), inside the second version of the formula, since the common difference is only added between terms, one fewer time than the number of terms.

It also helps to remember that an arithmetic series always has a constant difference between consecutive terms. If the ratio between terms is constant instead, you are looking at a geometric series, which uses a completely different sum formula.

When you meet an arithmetic series problem, work through three questions in order: What is the first term? What is the common difference (or the last term)? How many terms are being added? Once those three values are identified, the sum formula does the rest of the work for you.

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