# How do I find the sum of an arithmetic sequence on a calculator?

Mar 27, 2018

You can use the following general formula:

${\sum}_{k = 1}^{n} \left(a k + b\right) = \frac{n}{2} \left[a \left(n + 1\right) + 2 b\right]$

where the series has $n$ terms $a$ and $b$ are constants.

#### Explanation:

Seems that it would be handy to have a closed form expression for the following:

${\sum}_{k = 1}^{n} \left(a k + b\right)$

where a and b are constants.

The commutative property of addition says

${\sum}_{k = 1}^{n} \left(a k + b\right) = {\sum}_{k = 1}^{n} a k + {\sum}_{k = 1}^{n} b$

We can factor the $a$.

${\sum}_{k = 1}^{n} \left(a k + b\right) = a {\sum}_{k = 1}^{n} k + {\sum}_{k = 1}^{n} b$

But we know closed form representations for both of these sums.

${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$

and

${\sum}_{k = 1}^{n} b = n b$

So

${\sum}_{k = 1}^{n} \left(a k + b\right) = \frac{a n \left(n + 1\right)}{2} + n b = \frac{n}{2} \left[a \left(n + 1\right) + 2 b\right]$

EXAMPLE

Suppose we wanted the following sum:

$7 + 10 + 13 + 16 + 19 + 22 + 25 + 28 + 31$

There are 9 terms in this series so $n = 9$. Consecutive terms differ by 3 so $a = 3$. The first term, ${x}_{1} = 7$ and so $b = {x}_{1} - a = 7 - 3 = 4$.

We want to find

${\sum}_{k = 1}^{9} \left(3 k + 4\right) = \frac{9}{2} \left[3 \left(9 + 1\right) + 2 \cdot 4\right] = 171$

which is, in fact, the sum of the numbers given.