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Math 3

Lesson

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.

Study the pattern for the following sequence, and write down the next two terms.

$6$6, $2$2, $-2$−2, $-6$−6, $\editable{}$, $\editable{}$

Study the pattern for the following sequence:

$280,230,180,130,\ldots$280,230,180,130,…

State the common difference between consecutive terms.

A diving vessel descends below the surface of the water at a constant rate so that the depth of the vessel after $1$1 minute, $2$2 minutes and $3$3 minutes is $50$50 meters, $100$100 meters and $150$150 meters respectively.

By how much is the depth increasing each minute?

What will the depth of the vessel be after $4$4 minutes?

Continuing at this rate, what will be the depth of the vessel after $10$10 minutes?

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 $t_1=-3$`t`1=−3,

the second term is given by $t_2=5=-3+1\times8$`t`2=5=−3+1×8 ,

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

the fourth term $t_4=21=-3+3\times8$`t`4=21=−3+3×8 and so on.

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

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

In our example we have that $t_n=-3+\left(n-1\right)\times8$`t``n`=−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 $t_1=8\times1-11=-3$`t`1=8×1−11=−3 and that $t_2=8\times2-11=5$`t`2=8×2−11=5 and that $t_{100}=8\times100-11=789$`t`100=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 $t_n=87+\left(n-1\right)\times-7$`t``n`=87+(`n`−1)×−7

This simplifies to $t_n=94-7n$`t``n`=94−7`n` and we can use this formula to find any term we like. For example, the first negative term in the sequence is given by $t_{14}=94-7\times14=-4$`t`14=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 $t_5=38$`t`5=38 and $t_9=66$`t`9=66. This mean we can write down two equations:

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

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

If we now subtract equation (1) from equation (2) the first term in each equation will cancel out to leave us with $\left(8d-4d\right)=66-38$(8`d`−4`d`)=66−38. This means $4d=28$4`d`=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 $t_n=a+\left(n-1\right)d=10+\left(n-1\right)\times7$`t``n`=`a`+(`n`−1)`d`=10+(`n`−1)×7 and this simplifies to $t_n=3+7n$`t``n`=3+7`n`.

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

The $n$`n`th term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$`T``n`=15+5(`n`−1).

Determine $a$

`a`, the first term in the arithmetic progression.Determine $d$

`d`, the common difference.Determine $T_9$

`T`9, the $9$9th term in the sequence.

In an arithmetic progression where $a$`a` is the first term, and $d$`d` is the common difference, $T_7=43$`T`7=43 and $T_{14}=85$`T`14=85.

Determine $d$

`d`, the common difference.Determine $a$

`a`, the first term in the sequence.State the equation for $T_n$

`T``n`, the $n$`n`th term in the sequence.Hence find $T_{25}$

`T`25, the $25$25th term in the sequence.

We can define an arithmetic progression with what is known as a recursive formula. We begin by indicating a first term, say $t_1=a$`t`1=`a`, and then expressing the progression as:

$t_{n+1}=t_n+d,$`t``n`+1=`t``n`+`d`, with $t_1=a$`t`1=`a`

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

$t_2=t_1+\sqrt{2}=1+\sqrt{2}$`t`2=`t`1+√2=1+√2

$t_3=t_2+\sqrt{2}=\left(1+\sqrt{2}\right)+\sqrt{2}=1+2\sqrt{2}$`t`3=`t`2+√2=(1+√2)+√2=1+2√2

$t_4=t_3+\sqrt{2}=\left(1+2\sqrt{2}\right)+\sqrt{2}=1+3\sqrt{2}$`t`4=`t`3+√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 $t_{n+1}=t_n+n$`t``n`+1=`t``n`+`n`, with $t_1=0$`t`1=0. To test whether it is an arithmetic sequence, we will consider the first four terms. Here:

$t_1=0$`t`1=0,

$t_2=t_1+2=2$`t`2=`t`1+2=2,

$t_3=t_2+3=2+3=5$`t`3=`t`2+3=2+3=5,

$t_4=t_3+4=5+4=9$`t`4=`t`3+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.

Consider the first-order recurrence relationship defined by $T_n=T_{n-1}+2$`T``n`=`T``n`−1+2, $T_1=5$`T`1=5.

Determine the next four terms of the sequence, from $T_2$

`T`2 to $T_5$`T`5.Write all four terms on the same line, separated by commas.

Plot the first five terms on the graph below.

Loading Graph...Is the sequence generated from this definition arithmetic or geometric?

Arithmetic

AGeometric

BNeither

C

The first term of an arithmetic sequence is $2$2. The fifth term is $26$26.

Solve for $d$

`d`, the common difference of the sequence.Write a recursive rule for $T_n$

`T``n` in terms of $T_{n-1}$`T``n`−1 which defines this sequence and an initial condition for $T_1$`T`1.Write both parts on the same line separated by a comma.

Zuber is a taxi service that charges a $\$1.50$$1.50 pick-up fee and $\$1.95$$1.95 per kilometer of travel.

What is the total charge for a $10$10 km journey?

We want to describe this situation as a recursive sequence.

To start with, state the initial condition $T_0$

`T`0.Write a recurrence relation for $T_n$

`T``n` in terms of $T_{n-1}$`T``n`−1 which defines the price of a $n$`n`km trip.

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).

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 $t_n=2n+3$`t``n`=2`n`+3. We can cleverly rearrange this equation to $t_n=5+\left(2n-2\right)$`t``n`=5+(2`n`−2) . By taking out common factors on the last two terms, the expression becomes:

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

Comparing this to the general formula for the $n$`n`th term of an AP, given by $t_n=a+\left(n-1\right)d$`t``n`=`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 $t_n=dn+k$`t``n`=`d``n`+`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`=`m``x`+`c` is drawn as a straight line, so the arithmetic progression given by $t_n=dn+k$`t``n`=`d``n`+`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 $\left(t_1=5\right)$(`t`1=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.

The $n$`n`th term of an arithmetic progression is given by the equation $T_n=12+4\left(n-1\right)$`T``n`=12+4(`n`−1).

Complete the table of values.

$n$ `n`$1$1 $2$2 $3$3 $4$4 $10$10 $T_n$ `T``n`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ By how much are consecutive terms in the sequence increasing?

Plot the points in the table on the graph.

Loading Graph...If the points on the graph were joined, they would form:

a straight line

Aa curved line

B

The plotted points represent terms in an arithmetic sequence:

Loading Graph...

Complete the table of values for the given points.

$n$ `n`$1$1 $2$2 $3$3 $4$4 $T_n$ `T``n`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Identify $d$

`d`, the common difference between consecutive terms.Write a simplified expression for the general $n$

`n`th term of the sequence, $T_n$`T``n`.Find the $15$15th term of the sequence.

A racing car starts the race with $150$150 liters of fuel. From there, it uses fuel at a rate of $5$5 liters per minute.

What is the rate of change?

Fill in the table of values:

Number of minutes passed ($x$ `x`)$0$0 $5$5 $10$10 $15$15 Amount of fuel left in tank in liters ($y$ `y`)$150$150 $\editable{}$ $\editable{}$ $\editable{}$ Write an algebraic equation linking the number of minutes passed ($x$

`x`) and the amount of fuel left in the tank ($y$`y`).By rearranging the equation found in part (d), calculate how long it will take for the car to run out of fuel.

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 $n$`n`th term of an AP, given as $t_n=a+\left(n-1\right)d$`t``n`=`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 $t_{365}=365+24=389$`t`365=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 $S_n=\frac{n}{2}\left(a+l\right)$`S``n`=`n`2(`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.

A worker at a factory is stacking cylindrical-shaped pipes which are stacked in layers. Each layer contains one pipe less than the layer below it. There are $4$4 pipes in the topmost layer, $5$5 pipes in the next layer, and so on. There are $n$`n` layers in the stack.

Form an expression for the number of pipes in the bottom layer.

Show that there are a total of $n\left(\frac{n+7}{2}\right)$

`n`(`n`+72) pipes in the stack.

Bart is learning to drive. His first lesson is $26$26 minutes long, and each subsequent lesson is $4$4 minutes longer than the lesson before.

How long will his $15$15th lesson be?

If Bart reaches $9.6$9.6 total

*hours*of driving on his $n$`n`th lesson, solve for $n$`n`.

Katrina starts training for a $4.5$4.5 km charity trail run by running every week for $28$28 weeks.

She runs $2$2 km of the course in the first week, and each week after that she runs $250$250 meters more than the previous week.

She continues to increase the distance she runs until the week when she runs far enough to complete the course. Each week after that she completes the course without increase.

How far does she run in the $11$11th week? Give your answer correct to two decimal places, if necessary.

What is the total distance that Katrina runs in $28$28 weeks? Give your answer correct to two decimal places, if necessary.