The Sum of an Arithmetic Series

An arithmetic sequence of numbers, sometimes alternatively called an arithmetic progression, is a sequence of numbers in which the difference between all pairs of consecutive numbers is constant. A very simple arithmetic sequence consists of the natural numbers: 1, 2, 3, 4, … where the difference between any number and the number before it is just one. 3, 7, 11, 15, 19, …. is another arithmetic sequence, but in this case the constant difference between elements is four.

A finite portion of an arithmetic sequence like 2, 3, 4 or 7, 11, 15 is called a finite arithmetic progression. To confuse matters, sometimes a finite arithmetic progression, like an arithmetic sequence, is also called an arithmetic progression. To be safe, when a progression is finite, I always say as much.

An arithmetic series is the sum of a finite arithmetic progression. An arithmetic series consisting of the first four natural numbers is 1 + 2 + 3 + 4. The sum, 10, is trivial to compute via simple addition, but for a longer series with larger numbers, having a formula to calculate the sum is indispensable.

Consider a generic arithmetic series starting at a number $a_1$ and ending at a number $a_n$ . The series has a total of n terms with a constant difference d between each.

$a_1 + a_2 + a_3 + ... + a_{n-1} + a_n$

The formula to find the sum S of such a series is:

$S = \dfrac{n(a_1 + a_n)}{2}$

To see where this formula comes from and why the constant difference d doesn’t make an appearance in it (or does it?), let’s take a look at the derivation proof.

Proof

Given a generic arithmetic series starting at a number $a_1$ , ending at a number $a_n$ , with n terms and a constant difference d between each term, the sum S of the series is:

$S =a_1 + a_2 + a_3 + ... + a_{n-1} + a_n$

Since we know the difference d and number of terms n, we can rewrite the above equation as:

$S = a_1 + (a_1 + d) + (a_1 + 2d) + ... + [a_1 + (n-2)d] +[a_1 + (n-1)d]$

Now instead of writing the sum of the series in terms of the first term, let’s write in terms of the last term $a_n$ .

$S = [a_n - (n-1)d] + [a_n - (n-2)d] + ... + (a_n - 2d) + (a_n - d) + a_n$

Rearrange both of these equations collecting the lone $a_1$ s and $a_n$ s in each respectively.

$S = na_1 + d + 2d + ... + (n-2)d + (n-1)$
$S = na_n - (n-1)d - (n-2)d - ... - 2d - d$

If we add these two equations together, we see that all of the terms on the right side cancel out except for the first term in each equation.

$2S = na_1 + na_n$

Solve for S, and the proof is finished.

$S = \dfrac{n(a_1 + a_n)}{2}$

Random Thoughts

The sum of a generic arithmetic series can be written as a summation:

$\displaystyle\sum_{i=0}^{n-1}(a_1 + id) = \dfrac{n(a_1 + a_n)}{2}$

If we know the constant difference d and the number of terms n, we don’t need to know the last term $a_n$ . Instead we can write $a_n$ relative to $a_1$ :

$a_n = a_1 + (n - 1)d$

This lets us eliminate $a_n$ from the sum formula:

$\displaystyle\sum_{i=0}^{n-1}(a_1 + id) = \dfrac{n(a_1 +a_1 + (n - 1)d)}{2} = \dfrac{n(2a_1 + (n - 1)d)}{2}$

Alternatively, if we know the first term $a_1$ , the last term $a_n$ , and the difference between terms d then we can calculate the number of terms n in the series by:

$n = \dfrac{a_n - a_1}{d} + 1$

We can substitute this into our sum formula to eliminate n if we wanted:

$\displaystyle\sum_{i=0}^{n-1}(a_1 + id) = \frac{1}{2}\left(\dfrac{a_n - a_1}{d} + 1\right)(a_1 + a_n)$

Suppose that difference between terms in the arithmetic series was 1., ie., the terms are consecutive integers. We can eliminate d entirely from the last equation.

$\displaystyle\sum_{i=0}^{n-1}(a_1 + id) = \frac{1}{2}(a_n - a_1 + 1)(a_1 + a_n), \quad d = 1$

Since the difference between the terms is 1, it might be helpful to change the notation to illustrate that this is a simple summation over consecutive integers.

$\displaystyle\sum_{i=1}^{n}a_i = \displaystyle\sum_{i=0}^{n-1}(a_1 + i), \quad d = 1$

$\displaystyle\sum_{i=1}^{n}a_i = \frac{1}{2}(a_n - a_1 + 1)(a_1 + a_n), \quad d = 1$

Multiplying the polynomials on the right side and simplifying:

$\displaystyle\sum_{i=1}^{n}a_i = \frac{1}{2}(a_n^2 - a_1a_n + a_n - a_1^2 + a_1a_n +a_1)$

$\displaystyle\sum_{i=1}^{n}a_i = \frac{1}{2}(a_n^2 + a_n - a_1^2 + a_1)$

$\displaystyle\sum_{i=1}^{n}a_i = \frac{1}{2}(a_n(a_n + 1) - a_1(a_1 - 1))$

$\displaystyle\sum_{i=1}^{n}a_i = \dfrac{a_n(a_n + 1)}{2} - \dfrac{a_1(a_1 - 1)}{2}$

This last equation is the one often used to calculate the sum of an arithmetic series when the difference between terms is 1. This equation is especially nice when the first term $a_1$ is 0 or 1 because the subtrahend becomes 0 yielding:

$\displaystyle\sum_{i=1}^{n}a_i = \dfrac{a_n(a_n + 1)}{2}, \quad a_1 = 0 \quad or \quad a_1 = 1$

The Sum of an Arithmetic Series

Proof

Random Thoughts

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