What is an arithmetic sequence?

A sequence in which the difference between consecutive terms is constant. This constant difference is called the common difference and is usually denoted by d. Example: 1, 3, 5, 7, 9, ... is an arithmetic sequence with common difference d = 2.

What is the formula for the general term of an arithmetic sequence?

The n-th term of an arithmetic sequence with first term a and common difference d is a_n = a + (n-1)d. It is obtained by adding the common difference d to the first term (n-1) times.

What is the sum formula for an arithmetic sequence?

The sum of an arithmetic sequence with first term a, last term l, and n terms is S_n = n(a+l)/2. Substituting l = a + (n-1)d gives S_n = n{2a+(n-1)d}/2. It can be proved by reverse-order addition or by a staircase-to-rectangle geometric argument.

What is the arithmetic mean (in the context of a three-term sequence)?

If three numbers a, b, c form an arithmetic sequence in this order, then b is called the arithmetic mean of a and c. It satisfies b = (a+c)/2, i.e. b is the ordinary arithmetic average of a and c.

Chapter 1: Arithmetic Sequences

General term and sum formulas

Introduction

An arithmetic sequence is a sequence in which the difference between consecutive terms is constant. This constant difference is called the common difference.

Examples of arithmetic sequences

$1, 3, 5, 7, 9, \ldots$ (common difference $d = 2$)
$10, 7, 4, 1, -2, \ldots$ (common difference $d = -3$)
$3, 3, 3, 3, \ldots$ (common difference $d = 0$)

$a_1$

$a_2$

$a_3$

$a_4$

$\cdots$

$a_n$

$+d$

Figure 1: Each term of an arithmetic sequence increases by the common difference $d$.

Formula for the general term

Theorem (general term of an arithmetic sequence)

For an arithmetic sequence with first term $a_1 = a$ and common difference $d$, the $n$-th term is

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

Proof

By definition, the difference between consecutive terms is always $d$:

$$a_{k+1} - a_k = d \quad (\text{for every } k \geq 1)$$

To find $a_n$, write out the above equation for $k = 1, 2, \ldots, n-1$:

\begin{align} a_2 - a_1 &= d \\ a_3 - a_2 &= d \\ a_4 - a_3 &= d \\ &\vdots \\ a_n - a_{n-1} &= d \end{align}

Adding these term by term causes the intermediate terms to telescope:

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

Substituting $a_1 = a$ and rearranging yields

$$a_n = a + (n-1)d \quad \square$$

$a_1$

$a_2$

$a_3$

$a_n$

$a$

$a+d$

$a+2d$

$\cdots$

$a+(n-1)d$

add $(n-1)$ copies of $d$

Figure 2: Number-line representation of the general term $a_n = a + (n-1)d$.

Sum formula

Theorem (sum of an arithmetic sequence)

The sum of an arithmetic sequence with first term $a$, last term $l$, and $n$ terms is

$$S_n = \dfrac{n(a + l)}{2}$$

Equivalently, in terms of first term $a$, common difference $d$, and number of terms $n$:

$$S_n = \dfrac{n\{2a + (n-1)d\}}{2}$$

Proof 1: reverse-order addition

Write $S_n$ in both forward and reverse order:

\begin{align} S_n &= a_1 + a_2 + a_3 + \cdots + a_{n-1} + a_n \\ S_n &= a_n + a_{n-1} + a_{n-2} + \cdots + a_2 + a_1 \end{align}

Add the two equations term by term. In the $k$-th column, the forward $k$-th term $a_k$ lines up with the reverse $k$-th term $a_{n+1-k}$, and

$$a_k + a_{n+1-k} = \{a + (k-1)d\} + \{a + (n-k)d\} = 2a + (n-1)d = a_1 + a_n = a + l$$

which is the constant $a + l$, independent of $k$ (equal to the sum of the two endpoints). Since there are $n$ such columns,

$$2S_n = \underbrace{(a + l) + (a + l) + \cdots + (a + l)}_{n \text{ terms}} = n(a + l)$$

Therefore

$$S_n = \dfrac{n(a + l)}{2} \quad \square$$

Moreover, substituting $l = a_n = a + (n-1)d$ gives

$$S_n = \dfrac{n\{a + a + (n-1)d\}}{2} = \dfrac{n\{2a + (n-1)d\}}{2} \quad \square$$

which expresses the sum using only $a$, $d$, and $n$.

$S_n:$

$a_1$

$a_2$

$a_3$

$\cdots$

$a_{n-1}$

$a_n$

$S_n:$

$a_n$

$a_{n-1}$

$a_{n-2}$

$\cdots$

$a_2$

$a_1$

$2S_n:$

$a+l$

$\cdots$

$a+l$

$= n(a+l)$

Figure 3: Reverse-order addition — summing the forward and reverse rows gives $a+l$ in every column, so $2S_n = n(a+l)$.

Proof 2: geometric interpretation (area of a trapezoid)

Represent each term $a_k$ as a vertical bar of width $1$ and height $a_k$. Since each bar has area $a_k$, the total area of the bar chart equals

$$\text{staircase area} = a_1 + a_2 + \cdots + a_n = S_n$$

For an arithmetic sequence, the heights increase uniformly from $a$ to $l$, so the staircase has a trapezoidal profile. Flipping a second copy upside down and stacking it on top yields a rectangle whose height is the constant $a + l$ at every column:

$a_1$

$a_2$

$a_3$

$a_4$

$a_5$

$a$

$l$

$a+l$

$n$ terms

Figure 4: Combining two copies of the staircase gives a rectangle of width $n$ and height $a+l$. The area $= n(a+l) = 2S_n$, hence $S_n = \dfrac{n(a+l)}{2}$.

Arithmetic mean

Definition (arithmetic mean)

If three numbers $a, b, c$ form an arithmetic sequence in this order, then $b$ is called the arithmetic mean of $a$ and $c$.

Theorem

$a, b, c$ form an arithmetic sequence $\Leftrightarrow b = \dfrac{a + c}{2}$ ($b$ is the arithmetic mean of $a$ and $c$).

Proof

If $a, b, c$ form an arithmetic sequence with common difference $d$, then

$$b - a = d, \quad c - b = d$$

Hence $b - a = c - b$, which rearranges to

$$2b = a + c \quad \Leftrightarrow \quad b = \dfrac{a + c}{2} \quad \square$$

$a$

$b = \dfrac{a+c}{2}$

$c$

$d$

Figure 5: Three numbers $a, b, c$ form an arithmetic sequence $\Leftrightarrow$ $b$ is the arithmetic mean of $a$ and $c$.

Worked examples

Example 1

For the arithmetic sequence with first term $3$ and common difference $4$, find the $10$-th term and the sum of the first $10$ terms.

Solution

By the general-term formula,

$$a_{10} = 3 + (10-1) \cdot 4 = 3 + 36 = 39$$

By the sum formula,

$$S_{10} = \dfrac{10(3 + 39)}{2} = \dfrac{10 \cdot 42}{2} = 210$$

Example 2

An arithmetic sequence has $5$-th term $17$ and $12$-th term $45$. Find its first term and common difference.

Solution

Applying $a_n = a + (n-1)d$,

\begin{align} a_5 &= a + 4d = 17 \\ a_{12} &= a + 11d = 45 \end{align}

Subtracting the two equations gives $7d = 28$, so $d = 4$.

Substituting $d = 4$ into $a + 16 = 17$ gives $a = 1$.

Answer: first term $a = 1$, common difference $d = 4$.