Arithmetic Series

Learning goals

Read $\sum_{k=1}^{n} a_k$ as a sum with index and limits
Prove $S_n = \tfrac{n(a_1 + a_n)}{2}$ by reversing and adding
Use the $d$ form when the last term is unknown
Count $n - m + 1$ terms in a range
Derive $1 + 2 + \cdots + n = \tfrac{n(n+1)}{2}$

From a sequence to a series

A sequence is an ordered list of terms. A series is what you get when you add those terms together. The two ideas are close but not the same: the sequence $2, 5, 8, 11$ is a list, while the series $2 + 5 + 8 + 11$ is a single number, the sum $26$ .

When the terms come from an arithmetic sequence, their sum is called a finite arithmetic series. We write $S_n$ , the $n$ th partial sum, for the sum of the first $n$ terms:

S_n = a_1 + a_2 + a_3 + \cdots + a_n.

The subscript on $S_n$ counts how many terms are being added. So $S_5$ is the sum of the first five terms, and $S_{100}$ is the sum of the first hundred. The terms come from a sequence with a first term $a_1$ and a common difference $d$ . Because of that structure, we will be able to fold this whole sum into a short formula.

Summation notation

Writing out $a_1 + a_2 + \cdots + a_n$ every time is clumsy, so mathematicians pack it into a single symbol built from the Greek capital letter sigma, $\sum$ . The expression

\sum_{k=1}^{n} a_k

is read “the sum of $a_k$ as $k$ runs from $1$ to $n$ ,” and it is just shorthand for the same expanded sum you already know:

\sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n.

Every part of the symbol has a job:

The index is the letter $k$ , a counter that steps through the whole numbers one at a time.
The lower limit is the number below the sigma, here $1$ , which says where the counter starts.
The upper limit is the number above the sigma, here $n$ , which says where the counter stops.
The summand is the expression after the sigma, here $a_k$ , the thing being added, written once with the index inside it.

To turn a sigma back into an ordinary sum, you substitute each value of the index in turn and add the results. For instance, with the summand $2k + 1$ and the index running from $1$ to $4$ ,

\begin{aligned} \sum_{k=1}^{4} (2k + 1) &= (2\cdot 1 + 1) + (2\cdot 2 + 1) + (2\cdot 3 + 1) + (2\cdot 4 + 1) \\ &= 3 + 5 + 7 + 9 = 24. \end{aligned}

The letter chosen for the index does not matter, because it vanishes once the sum is written out: $\sum_{k=1}^{4}(2k+1)$ and $\sum_{i=1}^{4}(2i+1)$ are the very same number. A finite arithmetic series is simply this notation applied to an arithmetic sequence, with $a_k = a_1 + (k-1)d$ as the summand:

S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

Check your understanding

What is $\sum_{k=1}^{3} (3k - 2)$ ?

12

18

5

7

The reverse-and-add shortcut

Adding the terms one by one always works, but it is slow. The clever idea is to write the sum twice, once forward and once backward, and add the two copies together. Try it on a short series first, $2 + 5 + 8 + 11 + 14$ , by writing it above its own reverse and looking at what each column adds to.

Writing the series above its own reverse lines the terms up so every column totals the same first-plus-last value. Here the five columns each add to 16, so twice the sum is 5 times 16, which is 80, and the sum is 40.

Every column comes to the same total, $16$ , because the top row is climbing by $3$ each step while the bottom row is falling by that same $3$ . Whatever one side gains, the other side loses, so the two changes cancel and the column total never moves. That is true for any evenly spaced list, not only this one, so the same trick proves a formula that works for every arithmetic series.

Reverse and add: $2S_n = n(a_1 + a_n)$ #

Write the series out in order, from the first term to the last:

S_n = a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n.

Now write the very same sum a second time, but with the terms listed in reverse order, from the last term back to the first. The total is unchanged, because addition does not care about the order:

S_n = a_n + a_{n-1} + a_{n-2} + \cdots + a_2 + a_1.

Add the two lines column by column. The first column is $a_1 + a_n$ . Watch what happens as you move one column to the right. The top entry changes by $+d$ (from $a_1$ to $a_2$ ), while the bottom entry changes by $-d$ , the exact opposite (from $a_n$ to $a_{n-1}$ ). One change undoes the other, so the column total never changes, whether $d$ is positive or negative. Every single column adds to the same value, $a_1 + a_n$ .

There are $n$ columns, one for each term, so adding the two lines produces $n$ copies of $a_1 + a_n$ :

2 S_n = n\,(a_1 + a_n).

The left side is two copies of the series, so dividing both sides by $2$ isolates one copy:

S_n = \frac{n(a_1 + a_n)}{2}.

This formula has a reading worth remembering. Since $\frac{a_1 + a_n}{2}$ is the average of the first and last terms, the sum equals the number of terms times that average. Evenly spaced terms balance around their middle, so replacing all of them by their average and multiplying by how many there are gives the exact total.

A second form using the common difference

Sometimes you do not know the last term $a_n$ , only the first term $a_1$ , the common difference $d$ , and the number of terms $n$ . You can still find the sum by replacing $a_n$ with its formula from the previous lesson, $a_n = a_1 + (n-1)d$ :

S_n = \frac{n(a_1 + a_n)}{2} = \frac{n\big(a_1 + a_1 + (n-1)d\big)}{2} = \frac{n}{2}\big(2a_1 + (n-1)d\big).

This is the same total reached from a different starting point, so the two forms always agree; pick whichever matches the information you already have. Use $S_n = \frac{n(a_1 + a_n)}{2}$ when you know the first and last terms and how many there are. Use $S_n = \frac{n}{2}\big(2a_1 + (n-1)d\big)$ when you know the first term, the common difference, and the count, but not the last term.

Worked example 1 Sum the first 25 terms

Find the sum of the first $25$ terms of $3, 7, 11, 15, \ldots$

The first term is $a_1 = 3$ and the common difference is $d = 7 - 3 = 4$ . You know $a_1$ , $d$ , and $n = 25$ , but not the last term, so reach for the form that does not need $a_n$ :

S_{25} = \frac{25}{2}\big(2\cdot 3 + (25 - 1)\cdot 4\big) = \frac{25}{2}(6 + 96) = \frac{25}{2}\cdot 102.

Since $102 \div 2 = 51$ ,

S_{25} = 25 \times 51 = 1275.

You can confirm this with the other form by first finding the last term. The $25$ th term is $a_{25} = 3 + (25 - 1)\cdot 4 = 3 + 96 = 99$ , so

S_{25} = \frac{25(3 + 99)}{2} = \frac{25 \times 102}{2} = 1275.

Both routes give $1275$ , which is a good habit for catching slips.

Nothing about either formula requires $d$ to be positive. A decreasing sequence, where $d$ is negative, adds up the very same way.

Check your understanding

An arithmetic series has first term $a_1 = 50$ , common difference $d = -4$ , and $n = 20$ terms. What is $S_{20}$ ?

240

1{,}760

200

1{,}000

Worked example 2 The sum 1 + 2 + 3 + ... + 100

Add all the whole numbers from $1$ to $100$ .

This is an arithmetic series with first term $a_1 = 1$ , last term $a_{100} = 100$ , and $n = 100$ terms. Both the first and last terms are known, so use

S_{100} = \frac{n(a_1 + a_n)}{2} = \frac{100\,(1 + 100)}{2}.

The first and last terms add to $101$ , and there are $100$ of them, so

S_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050.

The pairing view says the same thing: match $1$ with $100$ , $2$ with $99$ , $3$ with $98$ , and so on down the line. That makes $50$ pairs, each adding to $101$ , and $50 \times 101 = 5050$ .

Nothing about that argument depends on stopping at $100$ . For any positive whole number $n$ , the sum $1 + 2 + \cdots + n$ is an arithmetic series with first term $a_1 = 1$ , last term $a_n = n$ , and $n$ terms, so

1 + 2 + \cdots + n = \frac{n(1 + n)}{2} = \frac{n(n+1)}{2}.

This single formula covers every whole-number sum at once, from $1 + 2 + \cdots + 100$ down to $1 + 2 + \cdots + 5$ , or up to any $n$ you choose.

Check your understanding

An arithmetic series has first term $a_1 = 5$ , last term $a_n = 95$ , and $n = 10$ terms. What is the sum $S_{10}$ ?

500

550

1{,}000

475

Summing a range of terms

Not every series starts at the first term. To add, say, the $10$ th through the $30$ th terms, notice that any run of consecutive terms of an arithmetic sequence is itself an arithmetic series. So the same formula applies, as long as you use that run’s own first term, its own last term, and the right count. The number of terms from position $m$ to position $n$ is $n - m + 1$ , not $n - m$ , because both endpoints are counted.

Worked example 3 Sum the 10th through 30th terms

For the arithmetic sequence with $a_1 = 4$ and $d = 3$ , find the sum of the terms from the $10$ th through the $30$ th.

Start with the explicit formula for a single term, $a_n = a_1 + (n-1)d = 4 + 3(n - 1)$ . The two end terms of the range are

a_{10} = 4 + 3(10 - 1) = 4 + 27 = 31, \qquad a_{30} = 4 + 3(30 - 1) = 4 + 87 = 91.

Count the terms carefully. From position $10$ to position $30$ there are $30 - 10 + 1 = 21$ terms, because both ends are included. This range is its own arithmetic series with first term $31$ , last term $91$ , and $21$ terms:

S = \frac{21(31 + 91)}{2} = \frac{21 \times 122}{2} = 21 \times 61 = 1281.

The sum of the terms from the $10$ th through the $30$ th is $1281$ .

Check your understanding

How many terms are there from the $7$ th term through the $19$ th term of a sequence, inclusive?

13

12

26

7

Series in the real world

Worked example 4 Seats in a section

A theater section has $12$ rows. The front row holds $18$ seats, and each row behind it holds $2$ more seats than the row in front. How many seats are in the whole section?

The seat counts by row form an arithmetic sequence with $a_1 = 18$ and $d = 2$ , and the section total is the arithmetic series of those $12$ row counts. First find the back row, the $12$ th term:

a_{12} = 18 + (12 - 1)\cdot 2 = 18 + 22 = 40.

The rows run from $18$ seats up to $40$ seats, and there are $12$ of them, so the total is

S_{12} = \frac{12(18 + 40)}{2} = \frac{12 \times 58}{2} = 6 \times 58 = 348 \text{ seats}.

Adding the twelve row counts one at a time would give the same $348$ , but the series formula reaches it in a single step. The very same method totals stacked logs, the rows of a stadium, or a saving plan that grows by a fixed amount each period.

Common mistakes

Confusing a sequence with a series. A sequence is the list $a_1, a_2, a_3, \ldots$ ; a series is its sum $a_1 + a_2 + a_3 + \cdots$ . “Find the $10$ th term” asks about the sequence, while “find the sum of the first $10$ terms” asks about the series.
Miscounting the terms in a range. The number of terms from position $m$ to position $n$ is $n - m + 1$ , not $n - m$ . Dropping the $+1$ loses one whole term: from the $10$ th to the $30$ th there are $21$ terms, not $20$ .
Forgetting to divide by two. The sum is $\frac{n(a_1 + a_n)}{2}$ , not $n(a_1 + a_n)$ . The division by $2$ is there because reverse-and-add counts the whole series twice before you halve it.
Choosing the less convenient form. Finding $a_n$ first and then using $S_n = \frac{n(a_1 + a_n)}{2}$ is not wrong when you know $a_1$ , $d$ , and $n$ , but it adds an extra step. Going straight to $S_n = \frac{n}{2}\big(2a_1 + (n-1)d\big)$ saves the trip.
Mixing up the count with the total. In “the sum of the first $12$ terms is $348$ ,” the number $12$ is the count $n$ and $348$ is the sum $S_n$ . Swapping them, treating the total as if it were the number of terms, is a common slip in word problems.

Key takeaways

A series is the sum of the terms of a sequence; the finite arithmetic series is $S_n = a_1 + a_2 + \cdots + a_n$ .
Summation notation writes it compactly: $\sum_{k=1}^{n} a_k$ means $a_1 + a_2 + \cdots + a_n$ , where $k$ is the index, $1$ and $n$ are the lower and upper limits, and $a_k$ is the summand.
Reverse and add proves $2S_n = n(a_1 + a_n)$ , so $S_n = \frac{n(a_1 + a_n)}{2}$ , the number of terms times the average of the first and last.
Substituting $a_n = a_1 + (n-1)d$ gives the equivalent form $S_n = \frac{n}{2}\big(2a_1 + (n-1)d\big)$ , which is handy when the last term is unknown.
The number of terms from position $m$ to position $n$ is $n - m + 1$ . That range is its own arithmetic series, so its sum is the count times the average of its own first and last terms.
The famous case $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ follows at once, since $a_1 = 1$ and $a_n = n$ .

Arithmetic Series

Learning goals

From a sequence to a series

Summation notation

The reverse-and-add shortcut

A second form using the common difference

Summing a range of terms

Series in the real world

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice