3.08 Further applications of growth and decay

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.

 

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

Question 2

 

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

Question 4

 

Applications of first order linear recurrence

The graph of a recurrence relation can provide a visual understanding of the way the sequence behaves. For instance it may show whether the terms are always increasing or always decreasing in size. It might also show that the terms change linearly, or as a quadratic curve, or as some other known form, and this might provide a hint as to a possible explicit form of the sequence. 

Consider for example a stockbroker's claim that  $\$200$$200 can be turned into a $\$1000$$1000 in $10$10 months. He claims that he can make $30%$30% a month on the investment, for the cost of $\$35$$35 a month. 

Such a claim seems incredulous, but given the percentage earnings and the monthly fee stated, a recurrence relation can be formed as:

$T_{n+1}=1.3T_n-35,T_1=200$Tn+1​=1.3Tn​−35,T1​=200

Note that the coefficient $1.3$1.3 is equivalent to $130%$130%, which is a $30%$30% increase on the previous month's amount as shown by the formula. The subtraction of $\$35$$35 represents the monthly fee. 

We can tabulate the size of the investment at the beginning of the month for the first $10$10 months as follows, using month numbers and the amounts rounded to whole dollars: 

Month	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6	$7$7	$8$8	$9$9	$10$10
$	$200$200	$225$225	$258$258	$300$300	$355$355	$426$426	$519$519	$640$640	$796$796	$1000$1000

The beginning of month $2$2 shows that the investment has grown to $\$225$$225, determined using the recurrence formula, so that $T_2=1.3\times200-35=225$T2​=1.3×200−35=225. Similarly, the beginning of month $3$3 is determined as $T_3=1.3\times225-35=257.50$T3​=1.3×225−35=257.50 which, rounded to whole dollars, becomes $\$258$$258.

The sequence of $10$10 months can be graphed using a 3-dimensional histogram as follows:

Note the increase in the original investment, with the rate of change in value increasing as well. Although not strictly geometric (the $\$35$$35 deduction affects the ratio between successive terms), the trend is certainly apparent - the higher the amount, the faster it grows. This type of growth is typical of recurrence relationships that have the form $T_{n+1}=k\times T_n\pm d$Tn+1​=k×Tn​±d with $k>1$k>1, and is an example of the first order linear recurrence relations we investigated in our last lesson.

 

 

Practice questions

Question 5