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

1. Is the sequence $-2,3,8,13,\cdots$ arithmetic?

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

2. Is the sequence $2,5,8,12,15,18,22$ arithmetic?

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

3. Determine the common difference of the arithmetic sequence $-1,-3,-5,-7,\cdots$.

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

4. What is the common difference of the arithmetic sequence $\frac{1}{2},\frac{3}{4},1,\frac{5}{4},\cdots$?

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

5. Write a formula for the $n$th term of the arithmetic sequence $4,7,10,13,\cdots$.

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

6. Find the $10$th term of an arithmetic sequence whose first term is $-3$ and whose common difference is $4$.

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 $33$.

7. Find the common difference of an arithmetic sequence whose first and fiftieth terms are $-1$ and $146$, respectively.

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 $3$.

8. How many terms are in the sequence $-5,-1,3,7,\cdots,75$?

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.

9. The third term of an arithemtic sequence is $15$, while the fifteenth term is $3$. Find the common difference and the first term.

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 $17$.

10. Find the arithmetic mean of $a$ and $b$.

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 $\frac{a+b}{2}$.

11. PROVE that three consecutive terms of an arithmetic sequence cannot add up to zero, unless one of the terms is zero.

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 $b=0$.

12. PROVE that three consecutive integers cannot add up to zero, unless one of the integers is zero.

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

13. Insert four arithmetic means between $3$ and $33$.

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.

14. 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.

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

1. 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.
2. Identify the first term and the common difference of each arithmetic sequence:
   • $-2,1,4,7,10,\cdots$
   • $\frac{-1}{4},\frac{1}{2},\frac{5}{4},2,\frac{11}{4},\cdots$
   • $0.8,-1.2,-3.2,-5.2,-7.2,\cdots$
3. The first term of an arithmetic sequence is $7$, while the common difference is $5$. Find the $75$th term of the sequence.
4. What is the $n$th term of the sequence $1,6,11,16,\cdots$?
5. Find the number of terms in the sequence $-2,1,4,\cdots,55$.
6. The houses on a particular street are assigned even numbers, starting from $2$. Find the number corresponding to the $100$th house.
7. PROVE that there is no term in the sequence $-3,1,5,\cdots$ that equals $67$.
8. 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 the slope between any two terms of an arithmetic sequence. Indeed, the graph of an arithmetic sequence is a straight line, whose slope corresponds to the common difference.
9. 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.
10. 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.