4.11 Arithmetics sequences as linear functions

An arithmetic sequence is a sequence in which the difference between successive terms is a constant. This constant is called the common difference and is usually denoted by the letter $d$d.

The progression $-3,5,13,21,\ldots$−3,5,13,21,… is an arithmetic progression with $a=-3$a=−3 and $d=8$d=8. On the other hand, the progression $1,10,100,1000,\ldots$1,10,100,1000,… is not arithmetic because the difference between each term is not constant.

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

 

Explicit formulas

An arithmetic progression starts with a first term, commonly denoted with the variable $a$a, and then either increases or decreases by a constant amount called the common difference $d$d. The progression $-3,5,13,21$−3,5,13,21 for example is an arithmetic progression with $a=-3$a=−3 and $d=8$d=8.

The generating rule for arithmetic progressions follows from this information. Using our example in the first paragraph,

we could say that the first term is given by $t_1=-3$t1​=−3,

the second term is given by $t_2=5=-3+1\times8$t2​=5=−3+1×8 ,

the third term is given by $t_3=13=-3+2\times8$t3​=13=−3+2×8 ,

the fourth term $t_4=21=-3+3\times8$t4​=21=−3+3×8 and so on.

 

The pattern starts to become clear and we could guess that the tenth term becomes $t_{10}=69=-3+9\times8$t10​=69=−3+9×8 and the one-hundredth term $t_{100}=789=-3+99\times8$t100​=789=−3+99×8

We could conclude that the generating rule for any arithmetic progression is given by $t_n=a+\left(n-1\right)d$tn​=a+(n−1)d where $t_n$tn​ is the $n$n^th term of the sequence.

In our example we have that $t_n=-3+\left(n-1\right)\times8$tn​=−3+(n−1)×8 and this can be simplified by distributing the parentheses and collecting like terms as follows:

$t_n$tn​	$=$=	$-3+\left(n-1\right)\times8$−3+(n−1)×8
$t_n$tn​	$=$=	$-3+8n-8$−3+8n−8
$t_n$tn​	$=$=	$8n-11$8n−11

Checking, we see that $t_1=8\times1-11=-3$t1​=8×1−11=−3 and that $t_2=8\times2-11=5$t2​=8×2−11=5 and that $t_{100}=8\times100-11=789$t100​=8×100−11=789.

Trying another example, the decreasing arithmetic sequence $87,80,73,66...$87,80,73,66..., ... has $a=87$a=87 and $d=-7$d=−7. Using our generating formula we have $t_n=87+\left(n-1\right)\times-7$tn​=87+(n−1)×−7

This simplifies to $t_n=94-7n$tn​=94−7n and we can use this formula to find any term we like. For example, the first negative term in the sequence is given by $t_{14}=94-7\times14=-4$t14​=94−7×14=−4.

Sometimes we are provided with two terms of an arithmetic sequence and then asked to find the generating rule. For example, suppose a certain arithmetic progression has $t_5=38$t5​=38 and $t_9=66$t9​=66. This mean we can write down two equations:

$a+4d=38$a+4d=38 (1)

$a+8d=66$a+8d=66 (2)

If we now subtract equation (1) from equation (2) the first term in each equation will cancel out to leave us with $\left(8d-4d\right)=66-38$(8d−4d)=66−38. This means $4d=28$4d=28 and so $d=7$d=7.

With the common difference found to be $7$7, then we know that, using equation (1) $a+4\times7=38$a+4×7=38 and so $a$a is clearly $10$10. The general term is given by $t_n=a+\left(n-1\right)d=10+\left(n-1\right)\times7$tn​=a+(n−1)d=10+(n−1)×7 and this simplifies to $t_n=3+7n$tn​=3+7n.

Checking, we see $t_5=3+7\times5=38$t5​=3+7×5=38 and $t_9=3+7\times9=66$t9​=3+7×9=66.

 

Practice questions

QUESTION 4

QUESTION 5

 

Recursive formulas

We can define an arithmetic progression with what is known as a recursive formula. We begin by indicating a first term, say $t_1=a$t1​=a, and then expressing the progression as:

$t_{n+1}=t_n+d,$tn+1​=tn​+d, with $t_1=a$t1​=a

So for example, the arithmetic progression given by the recursive equation $t_{n+1}=t_n+\sqrt{2}$tn+1​=tn​+√2 with $t_1=1$t1​=1 can be constructed term by term as:

$t_2=t_1+\sqrt{2}=1+\sqrt{2}$t2​=t1​+√2=1+√2

$t_3=t_2+\sqrt{2}=\left(1+\sqrt{2}\right)+\sqrt{2}=1+2\sqrt{2}$t3​=t2​+√2=(1+√2)+√2=1+2√2

$t_4=t_3+\sqrt{2}=\left(1+2\sqrt{2}\right)+\sqrt{2}=1+3\sqrt{2}$t4​=t3​+√2=(1+2√2)+√2=1+3√2

From the first four terms, it is clear that the sequence becomes:

$1,1+\sqrt{2},1+2\sqrt{2},1+3\sqrt{2},...,1+\left(n-1\right)\sqrt{2}+...$1,1+√2,1+2√2,1+3√2,...,1+(n−1)√2+...

An interesting sequence is given recursively as $t_{n+1}=t_n+n$tn+1​=tn​+n, with $t_1=0$t1​=0. To test whether it is an arithmetic sequence, we will consider the first four terms. Here:

$t_1=0$t1​=0,

$t_2=t_1+2=2$t2​=t1​+2=2,

$t_3=t_2+3=2+3=5$t3​=t2​+3=2+3=5,

$t_4=t_3+4=5+4=9$t4​=t3​+4=5+4=9,

The differences in successive terms is not constant, but rises by $1$1 each time, so there is a pattern to the sequence, but it is not arithmetic.

 

Practice questions

Question 6

Question 7

Question 8

 

Graphical representations

This applet will allow you to explore the geometrical features of an arithmetic progression, change the values of $a$a (the initial term) and $d$d (the common difference).

Guiding questions

1. In what ways is the formula for an arithmetic sequence similar to the equation of a line?

2. In what ways is the graph of an arithmetic sequence similar to the graph of a line?

Think about the generating rule for a particular arithmetic progression given by $t_n=2n+3$tn​=2n+3. We can cleverly rearrange this equation to $t_n=5+\left(2n-2\right)$tn​=5+(2n−2) . By taking out common factors on the last two terms, the expression becomes:

$t_n=5+\left(n-1\right)\times2$tn​=5+(n−1)×2

Comparing this to the general formula for the $n$nth term of an AP, given by $t_n=a+\left(n-1\right)d$tn​=a+(n−1)d , we can immediately see that the first term is $5$5 and the common difference is $2$2.

Thus any generating rule of the form $t_n=dn+k$tn​=dn+k where $d$d and $k$k are constants can be shown to be an arithmetic sequence. Just like the equation $y=mx+c$y=mx+c is drawn as a straight line, so the arithmetic progression given by $t_n=dn+k$tn​=dn+k is plotted as a series of points that are all in a straight line.

The first term is represented by the left-most point shown $\left(t_1=5\right)$(t1​=5) . The slope of the marked points measured, as the vertical distance between the points, is the common difference. In our example, the line of points is rising, so this indicates a positive common difference. In other instances, the line might be falling and this would indicate a negative common difference.

 

Practice questions

QUESTION 9

QUESTION 10

QUESTION 11

Practical applications

There is a range of everyday applications involving arithmetic sequences. If you are saving money in equal installments, for example, the cumulative savings at each savings period form an arithmetic sequence. If you are traveling down a highway at a constant speed, the amount of gasoline left in the tank, if measured every minute of the trip, forms another arithmetic progression. In fact any time you notice a quantity changing in equal amounts at set time periods, then you can consider that process as being arithmetic.

Tackling problems involving arithmetic sequences can often be made easier by generalizing the process by constructing a formula. Suppose for example a wealthy Chief Executive Officer decides to save an amount of $\$25$$25 birthday money given to her, but then adds to it according to the following plan. On the day after her birthday she will add a further $\$26$$26 to the savings. On day two, she will add another $\$27$$27 and again on day three she will add $\$28$$28, continuing on in this manner until the day before her next birthday. How much will the CEO have all together?

Such a problem could be answered by carefully adding up the installments day by day until the total is obtained. Not only is this time consuming, we might suspect our answers accuracy given the $365$365 additions required. Moreover, while the specific problem might be solved, we would need to repeat the procedure for any other similar problem. We will not have learned anything about the general features of the savings construction. Developing a formula on the other hand may reveal a type of generality that may be very useful in solving a range of similar problems.

Taking this general approach we might see the problem is arithmetic, and set $a=25$a=25 and $d=1$d=1 then using the formula for the $n$nth term of an AP, given as $t_n=a+\left(n-1\right)d$tn​=a+(n−1)d, we determine the rule for this particular process is

$t_n$tn​	$=$=	$25+\left(n-1\right)\times1$25+(n−1)×1
	$=$=	$n+24$n+24

The last term of the finite sequence corresponding to the amount saved on the day before her birthday is thus $t_{365}=365+24=389$t365​=365+24=389. This means that the sum of the savings sequence $25,26,27,...,389$25,26,27,...,389 can be determined using the the arithmetic sum formula $S_n=\frac{n}{2}\left(a+l\right)$Sn​=n2​(a+l) .

In our problem we determine specifically that

$S_{365}$S365​	$=$=	$\frac{365}{2}\left(25+389\right)$3652​(25+389)
	$=$=	$75555$75555

and so the total amount saved in the year is $\$75555$$75555.

 

Practice questions

QUESTION 12

QUESTION 13

QUESTION 14