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Concept Version 9
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Summing Terms in an Arithmetic Sequence

An arithmetic sequence which is finite has a specific formula for its sum.

Learning Objective

  • Calculate the sum of an arithmetic sequence up to a certain number of terms


Key Points

    • The sum of the members of a finite arithmetic sequence is called an arithmetic series.
    • The equation $\displaystyle{S_n = \left(\frac{n}{2} \right) \left [2a_1 + \left(n-1\right)d\right]}$ can be used to find the sum of any arithmetic sequence up to the $n$th term.
    • Some arithmetic sequences are infinite, and their general form is: $\displaystyle\sum_{i=1}^\infty {a_i} = \sum_{i=1}^\infty \left[a_1+\left(i-1\right)d\right]$. While these sequences are infinite, you can still apply the summation equation to find their sum up to a specific $n$th term.

Terms

  • infinite

    Boundless, endless, without end or limits; innumerable.

  • finite

    Limited, constrained by bounds.


Full Text

An arithmetic progression or arithmetic sequence is an ordered list of numbers such that the difference between the consecutive terms is constant. For instance, the sequence $5, 7, 9, 11, 13, \cdots$ is an arithmetic progression with common difference of $2$.

Finite Summation

The sum of the members of a finite arithmetic sequence is called an arithmetic series.

We can come up with a formula for the sum of a finite arithmetic formula by looking at the sum in two different ways. First we think of it as the sum of terms that are written in terms of $a_1$, so that the second term is $a_1+d$, the third is $a_1+2d$, and so on. Then our sum looks like:

$S_n = a_1 + \left(a_1 + d\right) + \left(a_1 +2d\right) + \cdots + \left[a_1 + \left(n-2\right)d\right] + \left[a_1 + \left(n-1\right)d\right]$

Next, we think of each term as being written in terms of the last term, $a_n$. Then the last term is $a_n$, the term before the last is $a_n-d$, the term before that is $a_n-2d$, and so on. The sum is therefore:

$S_n = \left[a_n - \left(n-1\right)d\right] + \left[a_n - \left(n-2\right)d\right] + \cdots + \left(a_n - 2d\right) + \left(a_n - d\right) + a_n$

Adding both sides of the two previous equations, all terms involving $d$ cancel and we obtain:

$2S_n = n\left(a_1 + a_n\right)$

Dividing both sides by $2$ produces a common form of the equation:

$\displaystyle{S_n = \left( \frac{n}{2} \right) \left (a_1 + a_n \right )}$

This can be thought of as $n$ times the average of the first and last terms. An alternate form results from reinserting the substitution $a_n = a_1 + \left(n-1\right)d$:

$\displaystyle{S_n = \left ( \frac{n}{2} \right ) \left[2a_1 + \left(n-1\right)d\right]}$

For example, consider the arithmetic progression $3,8,13,18,23, \cdots$ and suppose you want to know the sum up to the $50$th term. Note that you will need:

  • The number of terms you want to sum ($n$)
  • The first term of the sequence ($a_1$)
  • The difference in consecutive terms ($d$)

From the question, we know that $n=50$. We can see that the first term is $a_1 = 3$. The difference between the terms is $d = 5$. So the equation above gives: 

$\displaystyle{ \begin{aligned} S_{50} &=\left(\frac{50}{2}\right)\left(2 \cdot 3 + 49 \cdot 5\right) \\ &= 6275 \end{aligned} }$

Infinite Summation

An infinite arithmetic series is exactly what it sounds like: an infinite series whose terms are in an arithmetic sequence. Examples are $1 + 1 + 1 + 1 + \cdots$ and $1+2+3+4+ \cdots$. The general form for an infinite arithmetic series is:

$\displaystyle{\sum_{i=1}^\infty{a_i} = \sum_{i=1}^\infty{\left[a_1+\left(i-1\right)d\right]}}$

If $a_1=d=0$, then the sum of the series is $0$. If either $a_1$ or $d$ is non-zero, then the infinite series has no sum. 

Even if one is dealing with an infinite sequence, the sum of that sequence can still be found up to any $n$th term with the same equation used in a finite arithmetic sequence.

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