Arithmetic Sequences and Series, precalculus lesson

Seats in a theater that gain two in every row, a plant that grows the same amount each week, a stack of cans with one fewer can in each layer. Each list moves up or down by the same amount at every step. That one fixed step is enough to find any term at once and to add a long run of terms without adding them one by one.

Key terms

Common difference

Definition A sequence $\{a_n\}$ is arithmetic if there is a real number $d$ , called the common difference, such that

a_{n+1} - a_n = d

for every index $n$ for which $n + 1$ is also an index.

In plain words Every step adds the same number $d$ . If $d$ is negative the terms go down, and if $d = 0$ the sequence is constant.

Example In $7, 11, 15, 19, \dots$ every difference is $4$ , so the sequence is arithmetic with $d = 4$ . The list $20, 17, 14, 11$ is arithmetic with $d = -3$ . The list $2, 5, 9, 14$ is not arithmetic, because its differences $3, 4, 5$ are not all equal.

Explicit nth term

Definition The arithmetic sequence with first term $a_1$ and common difference $d$ has the explicit formula

a_n = a_1 + (n - 1)d \quad \text{for } n \ge 1.

Rewriting it as $a_n = dn + (a_1 - d)$ shows that $a_n$ is a linear function of $n$ with slope $d$ , so the points $(n, a_n)$ lie on a line. Conversely, every sequence $a_n = dn + b$ with constants $d$ and $b$ is arithmetic with common difference $d$ . This is the discrete version of a constant rate of change.

In plain words To get from the first term to the $n$ th term you take $n - 1$ steps, not $n$ , and each step has size $d$ .

Example For $7, 11, 15, \dots$ the formula is $a_n = 7 + 4(n - 1) = 4n + 3$ , so $a_{50} = 203$ without listing the terms in between. Two terms are also enough to find the sequence. If $a_5 = 17$ and $a_{12} = 45$ , then $12 - 5 = 7$ steps cover $45 - 17 = 28$ , so $d = 4$ and $a_1 = 17 - 4 \cdot 4 = 1$ .

Indexing

Definition The formula for the $n$ th term depends on where the indexing starts. If the first index is $0$ , then $a_n = a_0 + nd$ . For any two indices $m$ and $n$ of an arithmetic sequence,

a_n = a_m + (n - m)d.

An arithmetic list with first term $p$ , last term $q$ and common difference $d \ne 0$ has $N$ terms, where

N = \frac{q - p}{d} + 1.

In plain words Steps and terms are different counts. A fence with $10$ gaps has $11$ posts, and a list with $N$ terms has $N - 1$ steps between them.

Example The list $7, 11, 15, \dots, 203$ has $(203 - 7)/4 + 1 = 50$ terms. A plant is $12$ centimeters tall on the day it is planted and grows $1.5$ centimeters a week. Counting from $n = 0$ on planting day, its height after $n$ weeks is $h_n = 12 + 1.5n$ , so after $10$ weeks it is $h_{10} = 27$ centimeters tall. Using $a_1 + (n - 1)d$ here would give $25.5$ , the height after $9$ weeks, because that formula assumes the count starts at $1$ .

Pairing argument

Definition Let $S_n = a_1 + a_2 + \cdots + a_n$ for an arithmetic sequence. Write the sum forward and backward,

\begin{aligned} S_n &= a_1 + a_2 + \cdots + a_{n-1} + a_n, \\ S_n &= a_n + a_{n-1} + \cdots + a_2 + a_1. \end{aligned}

For each $k$ from $1$ to $n$ the two terms in column $k$ are $a_k$ and $a_{n+1-k}$ , and every column has the same total,

\begin{aligned} a_k + a_{n+1-k} &= \bigl(a_1 + (k - 1)d\bigr) \\ &\quad + \bigl(a_1 + (n - k)d\bigr) \\[0.4em] &= 2a_1 + (n - 1)d \\[0.4em] &= a_1 + a_n. \end{aligned}

Adding the two lines therefore gives $n$ equal columns, so $2S_n = n(a_1 + a_n)$ .

In plain words Pair the first term with the last, the second with the second to last, and so on. Every pair has the same total. Writing the sum twice avoids any trouble with an odd number of terms, where the middle term would have no partner.

Example In $1 + 2 + \cdots + 100$ the pairs $1 + 100$ , $2 + 99$ and so on up to $50 + 51$ each total $101$ . There are $50$ such pairs, so the sum is $50 \cdot 101 = 5050$ .

Arithmetic partial sums

Definition An arithmetic series is the sum of the terms of an arithmetic sequence. The $n$ th partial sum of the arithmetic sequence with first term $a_1$ and common difference $d$ is

\begin{aligned} S_n &= \sum_{k=1}^{n} \bigl(a_1 + (k - 1)d\bigr) \\[0.8em] &= \frac{n(a_1 + a_n)}{2} \\[0.8em] &= \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr). \end{aligned}

With $a_1 = 1$ and $d = 1$ this gives the sum of the first $n$ positive integers,

\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}.

An infinite arithmetic series diverges unless every term is $0$ , because otherwise its partial sums grow without bound in size.

In plain words Use the first form when you know the last term and the second form when you know the common difference.

Example The first $50$ terms of $7, 11, 15, \dots$ add up to

S_{50} = \frac{50(7 + 203)}{2} = 5250.

The sum rules give the same result. The numbers $1$ to $50$ add up to $50 \cdot 51 / 2 = 1275$ , so

\begin{aligned} \sum_{k=1}^{50} (4k + 3) &= 4 \cdot 1275 + 50 \cdot 3 \\[0.4em] &= 5250. \end{aligned}

Endpoint averaging

Definition For a finite arithmetic sequence, the average of all its terms equals the average of its first and last terms. So for any run of consecutive terms from $a_m$ to $a_n$ with $m \le n$ ,

\begin{aligned} &a_m + a_{m+1} + \cdots + a_n \\[0.4em] &\quad = (n - m + 1) \cdot \frac{a_m + a_n}{2}. \end{aligned}

When the number of terms is odd, this average is also the middle term.

In plain words Count the terms, average the two ends, and multiply. You never need the terms in between.

Example A theater has $20$ rows. The first row has $15$ seats and each row has $2$ more than the row in front of it. The last row has $15 + 19 \cdot 2 = 53$ seats, so the theater holds

20 \cdot \frac{15 + 53}{2} = 20 \cdot 34 = 680 \text{ seats.}

For an odd count, the sum $3 + 7 + 11 + 15 + 19$ has middle term $11$ , so it equals $5 \cdot 11 = 55$ .

Check yourself

An arithmetic sequence has $a_4 = 10$ and $a_9 = 30$ . Find $d$ , $a_1$ , a formula for $a_n$ and the term $a_{20}$ .

Show the answer

Five steps cover $30 - 10 = 20$ , so $d = 4$ . Then $a_1 = 10 - 3 \cdot 4 = -2$ and $a_n = -2 + 4(n - 1) = 4n - 6$ , which gives $a_{20} = 74$ .

Add the odd numbers from $1$ to $99$ .

Show the answer

There are $(99 - 1)/2 + 1 = 50$ terms, so the sum is

50 \cdot \frac{1 + 99}{2} = 2500.

Evaluate the sum below.

\sum_{k=5}^{24} (3k - 2)

Show the answer

The summand is linear in $k$ , so the terms form an arithmetic sequence. There are $24 - 5 + 1 = 20$ terms, the first is $13$ and the last is $70$ , so the sum is

20 \cdot \frac{13 + 70}{2} = 830.

Arithmetic Sequences and Series

Key terms

Common difference

Explicit nth term

Indexing

Pairing argument

Arithmetic partial sums

Endpoint averaging

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus