Arithmetic Sequences

One Fixed Step Between Consecutive Terms

A constant sequence repeats one value at every index. A natural next question is what happens when the terms do not stay fixed, but change by the same amount each time. The sequence $4, 7, 10, 13,\ldots$, for example, increases by $3$ at every step. The size and direction of this fixed change determine the sequence’s pattern.

Definition: A real sequence $(a_n)_{n=0}^{\infty}$ is an arithmetic sequence if there is a real number $d$ such that $$ a_{n+1}-a_n=d $$ for every $n\in\mathbb{N}_0$. The number $d$ is called the common difference.

The common difference is an amount added at each step. It can be positive, negative, or zero. If $d>0$, each step moves upward; if $d<0$, each step moves downward; and if $d=0$, every step leaves the value unchanged. Thus, constant sequences are exactly the arithmetic sequences with common difference zero.

When a formula is available, calculate $a_{n+1}-a_n$ and check whether the result is the same for every allowed index. For a sequence specified by its first term and common difference, the same information can instead be used to calculate any term directly. The following theorem makes that connection precise.

The Explicit Formula

Theorem: A sequence $(a_n)_{n=0}^{\infty}$ is arithmetic with common difference $d$ if and only if $$ a_n=a_0+nd $$ for every $n\in\mathbb{N}_0$. In particular, the first term $a_0$ and the common difference $d$ determine the entire sequence uniquely.

Proof. Suppose first that $(a_n)$ is arithmetic with common difference $d$. We prove by induction that $a_n=a_0+nd$ for every $n\in\mathbb{N}_0$. At $n=0$, the right-hand side is $a_0+0d=a_0$, so the statement holds. Now suppose it holds at some $n\in\mathbb{N}_0$. By the defining property of an arithmetic sequence,

a_{n+1}=a_n+d=(a_0+nd)+d=a_0+(n+1)d.

This proves the induction step. By the Principle of Mathematical Induction, the formula holds for every $n\in\mathbb{N}_0$.

Conversely, suppose $a_n=a_0+nd$ for every $n\in\mathbb{N}_0$. For every such $n$, substitution gives

a_{n+1}-a_n =\bigl(a_0+(n+1)d\bigr)-\bigl(a_0+nd\bigr) =d.

The difference between consecutive terms is therefore the same real number $d$ at every index, so the sequence is arithmetic with common difference $d$. Finally, any arithmetic sequence has $d=a_1-a_0$, so its common difference is unique. Its formula then determines every term from $a_0$ and $d$. $\square$

The formula can also be written relative to any chosen term. If $n\geq m$, then

$$ a_n=a_m+(n-m)d. $$

Indeed, substituting $a_m=a_0+md$ into the right-hand side gives $a_0+md+(n-m)d=a_0+nd=a_n$. This form is useful when a problem gives a term other than $a_0$.

Worked Example: Reading the Common Difference from a Formula

Define $p_n=11-4n$ for $n\in\mathbb{N}_0$. The explicit formula has the form $p_n=p_0+nd$, with $p_0=11$ and $d=-4$. To check the difference directly, calculate

p_{n+1}-p_n =\bigl(11-4(n+1)\bigr)-(11-4n) =11-4n-4-11+4n =-4.

Thus $(p_n)$ is arithmetic with common difference $-4$. Its first terms are $p_0=11$, $p_1=11-4=7$, and $p_2=11-8=3$; each consecutive subtraction gives $7-11=-4$ and $3-7=-4$, as expected.

Finding a Sequence from Given Terms

If a sequence is known to be arithmetic, two terms at different indices determine its common difference. The number of steps between the indices matters: subtracting the term values gives the total change across those steps, not necessarily the change in one step.

Worked Example: Using Terms at Different Indices

Suppose $(q_n)$ is arithmetic, $q_2=7$, and $q_6=-5$. From index $2$ to index $6$ there are four steps, so the total change $-5-7=-12$ is four times the common difference. Hence

4d=q_6-q_2=-5-7=-12, \qquad d=-3.

Using the formula relative to index $2$, for $n\geq2$,

$$ q_n=q_2+(n-2)d =7-3(n-2) =7-3n+6 =13-3n. $$

This formula also gives the earlier terms by using the full explicit formula: $q_0=13$ and $q_1=10$. Substitution checks both given values: $q_2=13-3(2)=13-6=7$, and $q_6=13-3(6)=13-18=-5$. The indices are nonnegative, so both checks are within the sequence’s domain.

The same reasoning works for any two indices $m<n$. If $(a_n)$ is arithmetic, then $a_n-a_m=(n-m)d$, so

d=\frac{a_n-a_m}{n-m}.

The condition $m<n$ ensures that the denominator is positive and, in particular, nonzero. A pair of terms can determine a candidate arithmetic sequence, but the assumption that the sequence is arithmetic is essential: two matching values alone do not prove that all intervening or later steps have a fixed difference.

Worked Example: Checking Whether a Formula Is Arithmetic

Consider $r_n=n^2+2$ for $n\in\mathbb{N}_0$. Its first terms are $r_0=2$, $r_1=3$, and $r_2=6$. The first two consecutive differences are

r_1-r_0=3-2=1, \qquad r_2-r_1=6-3=3.

Since $1\neq3$, the difference is not constant, so $(r_n)$ is not arithmetic. In fact, for every $n\in\mathbb{N}_0$,

r_{n+1}-r_n =\bigl((n+1)^2+2\bigr)-(n^2+2) =n^2+2n+1+2-n^2-2 =2n+1.

This difference depends on $n$, which confirms that it is not a common difference. The index values $n=0$ and $n=1$ give $1$ and $3$, respectively, agreeing with the direct calculations.

The Sign of the Common Difference

Theorem: Let $(a_n)$ be arithmetic with common difference $d$. If $d>0$, then the sequence is strictly increasing. If $d<0$, then it is strictly decreasing. If $d=0$, then it is constant.

Proof. By definition, $a_{n+1}-a_n=d$ for every $n\in\mathbb{N}_0$. If $d>0$, then $a_{n+1}-a_n>0$, which means $a_{n+1}>a_n$ at every index; this is precisely strict increase. If $d<0$, then $a_{n+1}-a_n<0$, so $a_{n+1}<a_n$ at every index; the sequence is strictly decreasing. If $d=0$, then $a_{n+1}=a_n$ for every $n$, so the earlier theorem on equal neighbors shows that the sequence is constant. $\square$

The conclusion works in the reverse direction too: if an arithmetic sequence is strictly increasing, then $a_1>a_0$, so $d=a_1-a_0>0$. If it is strictly decreasing, then $a_1<a_0$, so $d<0$. This makes the sign of $d$ a complete way to classify how an arithmetic sequence changes.

Worked Example: Interpreting a Negative Common Difference

Let $s_0=18$, and suppose $(s_n)$ is arithmetic with common difference $d=-\frac{5}{2}$. Its formula is

s_n=18-\frac{5}{2}n.

Because $d<0$, the theorem shows that the sequence is strictly decreasing. For instance, $s_1=18-\frac{5}{2}=\frac{31}{2}$, and $s_2=18-5=13$, with $s_2-s_1=13-\frac{31}{2}=-\frac{5}{2}$. This sequence does not consist only of integers; arithmetic sequences are allowed to have arbitrary real terms and common differences.

Combining Arithmetic Sequences

Arithmetic sequences also behave predictably under term-by-term addition and scalar multiplication. The change in a sum is the sum of the changes, and multiplying every term by a fixed number multiplies the common difference by that number.

Theorem: Suppose $(a_n)$ and $(b_n)$ are arithmetic sequences with common differences $d$ and $e$, respectively. For any real numbers $\alpha$ and $\beta$, the sequence $c_n=\alpha a_n+\beta b_n$ is arithmetic with common difference $\alpha d+\beta e$.

Proof. For every $n\in\mathbb{N}_0$, use the definitions of the two common differences to calculate

c_{n+1}-c_n =\bigl(\alpha a_{n+1}+\beta b_{n+1}\bigr) -\bigl(\alpha a_n+\beta b_n\bigr) =\alpha(a_{n+1}-a_n)+\beta(b_{n+1}-b_n) =\alpha d+\beta e.

This difference does not depend on $n$, so $(c_n)$ is arithmetic with common difference $\alpha d+\beta e$. $\square$

As a special case, the sum of two arithmetic sequences has common difference $d+e$, and their difference has common difference $d-e$. The result concerns addition and scalar multiplication; multiplying two arithmetic sequences term by term need not give an arithmetic sequence.

For example, in the sequence $(p_n)$ above with common difference $-4$, and the sequence $(q_n)$ with common difference $-3$, the term-by-term sum $p_n+q_n$ is arithmetic with common difference $-4+(-3)=-7$. This follows from the theorem, without needing to simplify the sum’s explicit formula.

Why the Fixed Difference Matters

An arithmetic sequence is controlled by just two pieces of information: its first term and its common difference. The explicit formula then gives any desired term, while the sign of the difference describes the direction of change. These facts are useful when translating between a recursive description, such as “add the same amount each time,” and a direct formula for the term at index $n$.

A common pitfall is to confuse the common difference with the difference between two terms whose indices are several steps apart. For example, $a_6-a_2$ is the accumulated change over four steps; the common difference is $(a_6-a_2)/4$, not $a_6-a_2$. Another pitfall is to assume that any sequence with two known terms must be arithmetic. The arithmetic property requires the same consecutive difference at every index.

Earlier in this course, the theorem on partial sums of an arithmetic sequence established how to sum a finite run of terms once the sequence has this form. Here the key step is recognizing and describing the sequence itself: determine its fixed difference, verify it does not depend on the index, and then use the explicit formula.

Check Your Understanding

Use the definition and the proved results to analyze the following sequences.

For $u_n=9+6n$, what are $u_0$ and the common difference? Is the sequence increasing, decreasing, or constant?
A sequence is arithmetic with $v_3=14$ and $v_7=2$. Find its common difference and an explicit formula for $v_n$.
Is $w_n=2n^2-1$ arithmetic? Compare consecutive terms to justify your answer.
If $(a_n)$ has common difference $5$ and $(b_n)$ has common difference $-2$, what is the common difference of $3a_n-2b_n$?
If an arithmetic sequence has common difference zero, what does the Equal Neighbors Theorem imply about the sequence?

What You'll Learn