Applications of geometric sequences
There are many applications of geometric sequences such as compound interest, exponential growth of bacteria, exponential decay of radioactive elements. If an amount is increasing or decreasing by a constant factor at set time periods, the process can be considered as being geometric.
A particular application of interest is the growth of an investment due to compound interest. You may recall from Chapter 2 that the compound interest formula is $A=P(1+r)^n$A=P(1+r)n. The form of this equation is very similar to the formula $u_n=aR^{n-1}$un=aRn−1.
The value of the investment after $n$n time periods is defined by $A$A. The principal $P$P, which is the initial value of the investment, is equivalent to $a$athe first term of a geometric sequence. The compound interest rate $r$r is expressed as a percentage and added to $1$1 to give the common ratio of a geometric sequence.
An amount of $\$2000$$2000 invested at a compound interest rate of $5%$5% per annum could be expressed as a geometric sequence with $a=2000$a=2000 and $r=1.05$r=1.05.
- If a quantity is increasing by $r%$r%, you must multiply by $(1+r%)$(1+r%) to find successive terms. The $1$1 represents $100%$100% of the original quantity and ensures that your calculation gives you the next term rather than just the amount by which it has increased.
- If a given interest rate in a question is compounding at a different frequency, you will need to change both the rate and the time periods. For example, 8% p.a. compounding quarterly for three years becomes a rate of $\frac{8}{4}=2%$84=2% per quarter and the number of time periods becomes $4\times3=12$4×3=12 quarters.
Worked example
Example 1
After receiving $\$500$$500 for her birthday, Hayley decides to spend $10%$10% 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$$405 will remain.
Do: The sequence of birthday dollars that remain at the beginning of the first five weeks is given as $500,450,405,364.5,328.05$500,450,405,364.5,328.05 and we can see here that this sequence is geometric with the first term $a=500$a=500 and the common ratio $r=0.9$r=0.9.
Hence, $B_n=500(0.9)^{n-1}$Bn=500(0.9)n−1.
(b) If instead of spending $10%$10%, Hayley decides to double her savings by setting aside an additional $10%$10% in savings each week. How long before she reaches her savings goal?
Think: Firstly, she will add $10%$10% to the birthday money, so that at the beginning of the second week, the additional $\$50$$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.
Do: The geometric progression for this plan becomes $500,550,605,665.5,732.05,...$500,550,605,665.5,732.05,...
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 $u_n=500\left(1.1\right)^{n-1}$un=500(1.1)n−1.
To solve for when the balance reaches her goal of $\$1000$$1000, solve for $n$n when $u_n=1000$un=1000;
| $500\left(1.1\right)^{n-1}$500(1.1)n−1 | $=$= | $1000$1000 | |
| $1.1^{n-1}$1.1n−1 | $=$= | $2$2 | Dividing both sides by 2 |
| $\therefore(n-1)$∴(n−1) | $\approx$≈ | $7.27$7.27 | Solve by guess and check, technology or logarithms. |
| $n$n | $\approx$≈ | $8.27$8.27 |
So $n$n must be greater than $8.27$8.27 for Hayley to have achieved her goal, so at the beginning of the $9$9th week her savings will exceed $\$1000$$1000.
The initial condition $u_1$u1 or $u_0$u0?
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 $u_1$u1 for the initial condition, referring to the first term of the sequence. However, sometimes it can be useful to use $u_0$u0 meaning the initial term. $u_0$u0 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 $u_0=100$u0=100 then $5$5 weeks later is $u_5$u5 where as if we used the initial condition of $u_1=100$u1=100 then $5$5 weeks later is $u_6$u6, which can be a bit confusing. Both would give the same answer but using $u_0$u0 in this case makes the term number we are looking for match the number of weeks.
Practice questions
QUESTION 1
The average rate of depreciation of the value of a Ferrari is $14%$14% per year. A new Ferrari is bought for $\$90000$$90000.
What is the car worth after $1$1 year?
What is the car worth after $3$3 years?
Write a recursive rule for $V_n$Vn defining the value of the car after $n$n years, and an initial condition $V_0$V0.
Write both parts of the rule on the same line, separated by a comma.
QUESTION 2
The first blow of a hammer drives a post a distance of $144$144 cm into the ground. Each successive blow drives the post $\frac{5}{6}$56 as far as the preceding blow. In order for the post to become stable, it needs to be driven $\frac{4651}{9}$46519 cm into the ground.
If $n$n is the number of hammer strikes needed for the pole to become stable, find $n$n.