Summary

An *arithmetic* sequence is a sequence in which consecutive terms increase (or decrease) by a *constant* number.
The constant number is called the *common difference* of the arithmetic sequence.

The general term of an *arithmetic sequence* is given by $T_{n}=a+(n-1)d$. Here:

- $a$ is called the first term;
- $d$ is called the common difference;
- $n$ is the number of terms;
- $T_{n}$ is the nth term.

There are *four* variables in the formula above: $n,T_{n},a,d$. You might be asked to find any of these, meaning that you should be able to rearrange the formula depending on the demands of a given question.

Examples

- Is the sequence $-2,3,8,13,\cdots$ arithmetic?
- Is the sequence $2,5,8,12,15,18,22$ arithmetic?
- Determine the common difference of the arithmetic sequence $-1,-3,-5,-7,\cdots$.
- What is the common difference of the arithmetic sequence $\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\cdots$?
- Write a formula for the $n$th term of the arithmetic sequence $4,7,10,13,\cdots$.
- Find the $10$th term of an arithmetic sequence whose first term is $-3$ and whose common difference is $4$.
- Find the common difference of an arithmetic sequence whose first and fiftieth terms are $-1$ and $146$, respectively.
- How many terms are in the sequence $-5,-1,3,7,\cdots,75$?
- The third term of an arithemtic sequence is $15$, while the fifteenth term is $3$. Find the common difference and the first term.
- Find the
*arithmetic mean*of $a$ and $b$. -
**PROVE**that*three consecutive terms of an arithmetic sequence cannot add up to zero, unless one of the terms is zero.* -
**PROVE**that*three consecutive integers cannot add up to zero, unless one of the integers is zero.* - Insert four
*arithmetic means*between $3$ and $33$. - Let $a$ and $z$ be positive integers, with $z>a$.
**PROVE**that the maximum number of arithmetic means that can be inserted between $a$ and $z$ is $z-a-1$, provided each arithmetic mean is an integer.

Yes, since $3-(-2)=8-3=13-8=5$, the terms increase by a constant number, namely $5$. Therefore, the sequence is arithmetic.

No. The difference between consecutive terms is *not* the *same*. For example, $12-8=4$, but $15-12=3$.

Take any two consecutive terms, e.g., the first two. We have $d=-3-(-1)=-2$. So, the common difference is $-2$.

Here, $d=\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{1\times 2}{2\times 2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}$.

$T_{n}=a+(n-1)d=4+(n-1)3=3n+1$

Here, we are given that $a=-3,~d=4$. We need the $10$th term, so $n=10$. Using $T_{n}=a+(n-1)d$ gives $$T_{10}=-3+(10-1)4=-3+9\times 4=33.$$ The $10$th term of the sequence is $33$.

Here, $a=-1$ and $T_{50}=146$. We need $d$. Substituting into the formula for the $n$th term gives $$146=-1+(50-1)d.$$ After combining like terms, the equation reduces to $147=49d$, whence $d=\frac{147}{49}=3$. So, the common difference is $3$.

We have $a=-5,d=4$ and $T_{n}=75$. We need $n$. We substitute the givens in the formula $T_{n}=a+(n-1)d$: $$75=-5+(n-1)4,$$ and obtain $n=21$. So there are $21$ terms in the sequence.

We have $T_{3}=15$ and $T_{15}=3$. Using $T_{n}=a+(n-1)d$, the resulting equations are $$a+2d=15$$ and $$a+14d=3.$$ The above linear system can be solved by elimination to obtain $d=-1$ and $a=17$. So, the common difference is $-1$ while the first term is $17$.

We need a third number $m$ that makes the sequence $a,m,b$ arithmetic. Since the common difference should be the same, we have $$m-a=b-m,$$ so $2m=a+b$, whence $m=\frac{a+b}{2}$. The desired arithmetic mean is $\frac{a+b}{2}$.

Suppose $a,b,c$ is an arithmetic sequence in which $a+b+c=0$. In view of the previous example, we have $2b=a+c$. Thus, $$0=a+b+c=b+a+c=b+2b=3b,$$ and so $b=0$.

Consecutive integers form an arithmetic sequence in which the common difference is $1$. So, this is a particular instance of the previous example.

Let the arithmetic means be $r,s,t,u$. Then the sequence $$3,r,s,t,u,33$$ is arithmetic. Observe that there are now $6$ terms in the sequence; in particular, $T_{6}=33$. So, $$33=3+(6-1)d$$ from which $d=6$. Thus, $r=9,s=15,t=21,u=27$. These are the desired arithmetic means.

Let the number of arithmetic means be $k$. We can enumerate the entire arithmetic sequence as $a,m_{1},m_{2},\cdots,m_{k},z$, where the $m_{i}$'s are the arithmetic means. Since the arithmetic sequence now contains $(k+2)$ terms and the $(k+2)$-th term is $z$, the use of the $n$th term formula gives $$z=a+(k+2-1)d.$$ In order to have all terms integers, and to have the maximum possible number of terms, the value of $d$ must be $1$. So, $z=a+(k+2-1)\times 1$. Now isolate $k$ to obtain $k=z-a-1$, as desired.

Class exercises

- Consider a degenerate case in which the common difference of an arithmetic sequence is
*zero*. Describe the general appearance of the arithmetic sequence in this case. - Identify the first term and the common difference of each arithmetic sequence:
- $-2,1,4,7,10,\cdots$
- $\frac{-1}{4},\frac{1}{2},\frac{5}{4},2,\frac{11}{4},\cdots$
- $0.8,-1.2,-3.2,-5.2,-7.2,\cdots$

- The first term of an arithmetic sequence is $7$, while the common difference is $5$. Find the $75$th term of the sequence.
- What is the $n$th term of the sequence $1,6,11,16,\cdots$?
- Find the number of terms in the sequence $-2,1,4,\cdots,55$.
- The houses on a particular street are assigned
*even*numbers, starting from $2$. Find the number corresponding to the $100$th house. -
**PROVE**that there is no term in the sequence $-3,1,5,\cdots$ that equals $67$. - Suppose that two terms of an arithmetic sequence are given as $T_{n}=R$ and $T_{m}=S$.
**PROVE**that the*common difference*$d$ can be given by $d=\frac{R-S}{n-m}$.

It follows that one can think of the*common difference*as thebetween any two terms of an arithmetic sequence. Indeed, theslope

graph

of an arithmetic sequence is a straightline

, whose slope corresponds to the common difference. - Consider an arithmetic sequence with at least two terms.
**PROVE**that $T_{n}=d$ for some $n\geq 2$*if, and only if,*$T_{m}=0$ for some $m~ (=n-1)$.

In other words, if an arithmetic sequence contains a $0$, then every other term is a multiple of the common difference. **PROVE**that if $T_{n}=a$ for an arithmetic sequence, then either the sequence is a constant sequence, or the sequence contains only one term.

Overall expectations

- C1: demonstrate an understanding of recursive sequences, represent recursive sequences in a variety of ways, and make connections to Pascalâ€™s triangle;
- C2: demonstrate an understanding of the relationships involved in arithmetic and geometric sequences and series, and solve related problems;
- C3: make connections between sequences, series, and financial applications, and solve problems involving compound interest and ordinary annuities.

Specific expectations

- C1.1 make connections between sequences and discrete functions, represent sequences using function notation, and distinguish between a discrete function and a continuous function [e.g., $f(x) = 2x$, where the domain is the set of natural numbers, is a discrete linear function and its graph is a set of equally spaced points; $f(x) = 2x$, where the domain is the set of real numbers, is a continuous linear function and its graph is a straight line]
- C1.3 connect the formula for the nth term of a sequence to the representation in function notation, and write terms of a sequence given one of these representations or a recursion formula

Learning goal

Success criteria