NZ Level 7 (NZC) Level 2 (NCEA) Applications of Arithmetic Progressions
Lesson

There are a range of every day applications involving arithmetic sequences. If you are saving money in equal instalments for example, the cumulative savings at each savings period form an arithmetic sequence. If you are travelling down a highway at a constant speed, the amount of petrol 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 generalising 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 instalments 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 learnt 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 $a=250$a=250 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+(n1)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$$75555. We could ask a different question about the specific savings plan the CEO has constructed. We might for example ask how long it would take for the savings to reach \10100$$10100. That is, we are asking for what specific value of $n$n will $S_n=10100$Sn=10100. To answer this recall that the sum of an arithmetic progression can also be expressed as $S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$Sn=n2[2a+(n1)d]. Putting in the specific parameters for this problem, we see that

 $S_n$Sn​ $=$= $\frac{n}{2}\left[2\times50+\left(n-1\right)\times1\right]$n2​[2×50+(n−1)×1] $=$= $\frac{n}{2}\left(99+n\right)$n2​(99+n)

Thus we can solve the problem by setting $\frac{n}{2}\left(99+n\right)=10100$n2(99+n)=10100. This is a quadratic equation and could be solved by rearranging first and then using factorisation techniques. However if we look carefully at the equation, we might see that $n=101$n=101 days is an obvious solution. In other instances we may not be so lucky.

Also note that our formulae provides the opportunity to find intermediate savings states. For example, $99$99 days on from the CEO's birthday, we can easily determine that $t_{100}=124$t100=124 and thus

 $S_{100}$S100​ $=$= $\frac{100}{2}\left(25+149\right)$1002​(25+149) $=$= $8700$8700

or $\$87008700. Also note that any other arithmetic savings plan can be interrogated in the same way.

#### Worked Examples

##### QUESTION 1

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.

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

2. Show that there are a total of $n\left(\frac{n+7}{2}\right)$n(n+72) pipes in the stack.

##### QUESTION 2

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.

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

2. If Bart reaches $9.6$9.6 total hours of driving on his $n$nth lesson, solve for $n$n.

##### QUESTION 3

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 metres 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.

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

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

### Outcomes

#### M7-3

Use arithmetic and geometric sequences and series

#### 91258

Apply sequences and series in solving problems