Applications of arithmetic sequences
We have seen many applications of linear growth or decay when studying linear functions previously. Hence, arithmetic sequences apply in many areas of life, including simple interest earnings, straight-line depreciation, monthly rental accumulation and many others. For example, if you are saving money in equal instalments, 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.
Depreciation
Depreciation refers to the situation where items or investments lose value over time. We will consider two types of depreciation in this course:
- Straight line depreciation - the value reduces by the same amount every time period. This is an application of arithmetic sequences.
- Reducing balance depreciation - the value reduces by a percentage of the previous value every time period. This is an application of geometric sequences.
This application is will be examined in further detail in the financial applications section of this course.
Worked example
Example 1
Tabitha starts with $\$200$$200 in her piggy bank, the following week she adds $\$25$$25 and then continues to add $\$25$$25 at the start of each successive week.
(a) Find a rule to describe $B_n$Bn the balance of her savings at the start of each week.
Think: The sequence of savings generated is $\$200,\$225,\$250,\$275\dots$$200,$225,$250,$275… This is arithmetic, so write down the starting value, $a$a, and common difference, $d$d, and use $t_n=a+\left(n-1\right)d$tn=a+(n−1)d to find a general rule.
Do: $a=200$a=200 and $d=25$d=25 and so our general rule is: $B_n=200+25(n-1)$Bn=200+25(n−1) or equivalently $B_n=175+25n$Bn=175+25n.
(b) Find when her savings will reach $450.
Think: To find when the savings reach $\$450$$450, we substitute this into our general rule and solve for $n$n.
Do:
| $450$450 | $=$= | $175+25n$175+25n |
| $\therefore25n$∴25n | $=$= | $275$275 |
| $n$n | $=$= | $11$11 |
Hence, at the start of the $11$11th week her savings will have grown to $\$450$$450.
The initial condition $t_1$t1 or $t_0$t0?
When we describe a recursive rule we know that it requires two parts. Firstly the rule describing how the sequence recurs and secondly, the initial condition, describing where to start. In previous chapters we have mainly used $t_1$t1 for the initial condition, referring to the first term of the sequence. However, sometimes it can be useful to use $t_0$t0 meaning the initial term. $t_0$t0 is particularly useful in financial questions and population questions where we start with an initial amount and then look for the amount in the days/weeks/month after that starting point. For example, consider a situation where we start with $\$100$$100 and each week this amount increases by $\$20$$20. If we want to know how much we have after $5$5 weeks and we use the initial condition of $t_0=100$t0=100 then $5$5 weeks later is $t_5$t5 where as if we used the initial condition of $t_1=100$t1=100 then $5$5 weeks later is $t_6$t6, which can be a bit confusing. Both would give the same answer but using $t_0$t0 in this case makes the term number we are looking for match the number of weeks.
Practice questions
Question 1
Zuber is a taxi service that charges a $\$1.50$$1.50 pick-up fee and $\$1.95$$1.95 per kilometre 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 $T_0$T0.
Write a recurrence relation for $T_n$Tn in terms of $T_{n-1}$Tn−1 which defines the price of a $n$n km trip.
Question 2
A car bought at the beginning of 2009 is worth $\$1500$$1500 at the beginning of 2015. The value of the car has depreciated by a constant amount of $\$50$$50 each year since it was purchased.
What was the car purchased for in 2009?
Plot the value of the car, $V_n$Vn, on the graph from 2009 (represented by $n=0$n=0) to 2015 (represented by $n=6$n=6).
Loading Graph...Write an explicit rule for the value of the car after $n$n years.
Give the rule in its expanded form.
Solve for the year $n$n at the end of which the car will be worth half the price it was bought for.
Applications of geometric sequences
It is also very common to encounter applications of geometric sequences such as compound interest, exponential growth of bacteria, exponential decay of radioactive elements. If we have an amount increasing or decreasing by a constant factor at set time periods, then you can consider that process as being geometric.
Worked example
Example 2
After receiving $\$500$$500 for her birthday, Hayley decides to spend $10%$10% of what remains of this money each week.
(a) Find a model for $B_n$Bn, the amount of birthday money she has left at the start of the $n$nth week.
Think: At the beginning of the second week, Hayley will have spent $\$50$$50, and will only have $\$450$$450 or $90%$90% of the birthday money left. During the second week she will spend slightly less - $10%$10% of her remaining $\$450$$450 or $\$45$$45, so at the beginning of the third week $\$405$$405 will remain. The sequence of birthday dollars that remain at the beginning of the first five weeks is given as $500,450,405,364.5,328.05$500,450,405,364.5,328.05 and we can see here that this sequence is geometric with the first term $a=500$a=500 and the common ratio $r=0.9$r=0.9.
Do: Using the explicit rule $t_n=a(r)^{n-1}$tn=a(r)n−1 we get, $B_n=500(0.9)^{n-1}$Bn=500(0.9)n−1.
(b) If instead of spending $10%$10%, Hayley decides to double her savings by setting aside an additional $10%$10% in savings each week. How long before she reaches her savings goal?
Think: Firstly, she will add $10%$10% to the birthday money, so that at the beginning of the second week, the additional $\$50$$50 will take the total to $\$550$$550. At the beginning of week $3$3 she will add a further $10%$10% of this accrued $\$550$$550, so that the new total becomes $\$605$$605. She will keep adding $10%$10% of what ever is there at the beginning of each week until she reaches or exceeds $\$1000$$1000.
The geometric progression for this plan becomes $500,550,605,665.5,732.05,...$500,550,605,665.5,732.05,...
Do: This progression has $a=500$a=500 and $r=\frac{550}{500}=1.1$r=550500=1.1, so that the $n$nth term of the progression is given by $t_n=500\left(1.1\right)^{n-1}$tn=500(1.1)n−1.
To solve for $n$n we can either solve the equation using the solve feature on our calculators or generate the terms of the sequence in the sequence and look for the first value to exceed her goal of $\$1000$$1000.
To solve for when the balance reaches her goal of $\$1000$$1000, solve for $n$n when $t_n=1000$tn=1000:
| $500\left(1.1\right)^{n-1}$500(1.1)n−1 | $=$= | $1000$1000 |
| $n$n | $\approx$≈ | $8.27$8.27 |
So $n$n must be greater than $8.27$8.27 for Hayley to have achieved her goal, so at the beginning of the $9$9th week her savings will exceed $\$1000$$1000.
Practice questions
Question 3
Suppose you save $\$1$$1 the first day of a month, $\$2$$2 the second day, $\$4$$4 the third day, $\$8$$8 the fourth day, and so on. That is, each day you save twice as much as you did the day before.
What will you put aside for savings on the $17$17th day of the month?
What will you put aside for savings on the $29$29th day of the month?
Question 4
To test the effectiveness of a new antibiotic, a certain bacteria is introduced to a body and the number of bacteria is monitored. Initially, there are $14$14 bacteria in the body, and after four hours the number is found to quadruple.
If the bacterial population continues to quadruple every four hours, how many bacteria will there be in the body after 24 hours?
The antibiotic is applied after 24 hours, and is found to kill half of the germs every four hours. How many bacteria will there be left in the body 24 hours after applying the antibiotic?
Assume the bacteria stops multiplying and round your answer to the nearest integer if necessary.