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5. Number Patterns and Recursions
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VCE 11 General 2023

5.06 Applications of geometric sequences

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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 4  that the compound interest formula is A=P(1+r)^n. The form of this equation is very similar to the formula u_n= aR^{n - 1}.

The value of the investment after n time periods is defined by A . The principal P, which is the initial value of the investment, is equivalent to a the first term of a geometric sequence. The compound interest rate r is expressed as a percentage and added to 1 to give the common ratio of a geometric sequence.

An amount of \$2000 invested at a compound interest rate of 5\% per annum could be expressed as a geometric sequence with a=2000 and r=1.05.

Some conditions to note on compound interest:

  • If a quantity is increasing by r\%, you must multiply by (1+r\%) to find successive terms. The 1 represents 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\% \text{ p.a.} compounding quarterly for three years becomes a rate of \dfrac{8}{4}=2\% per quarter and the number of time periods becomes 4\times3=12 quarters.

Examples

Example 1

After receiving \$500 for her birthday, Hayley decides to spend 10\% of this money each week.

a

Find a model for B_n, the amount of birthday money she has left at the start of the nth week.

Worked Solution
Create a strategy

Find R to substitute into u_n=aR^{n-1}.

Apply the idea

If 10\%=0.10 of her money is spent each week, then she is left with 1-0.1=0.9 of money each week. This means R=0.9.

We are also given with a=500 and u=B to substitute into the formula.

\displaystyle u_n\displaystyle =\displaystyle aR^{n-1}Write the formula
\displaystyle B_n\displaystyle =\displaystyle 500 \times 0.9^{n-1}Substitute u=B, a= 500, R= 0.9
b

If instead of spending 10\%, Hayley decides to double her savings by setting aside an additional 10\% in savings each week. How long before she reaches her savings goal?

Worked Solution
Create a strategy

Substitute the goal savings into B_n and new R value into B_n=500 \times R^{n-1}.

Apply the idea

The goal savings is double of her initial money so B_n=500 \times 2 = 1000.

If 10\%=0.10 of her money is added each week, then she will gain 1+0.1=1.1 of money each week. This means R=1.1.

\displaystyle B_n\displaystyle =\displaystyle 500 \times R^{n-1}Write the formula
\displaystyle 1000\displaystyle =\displaystyle 500 \times 1.1^{n-1}Substitute B_n=1000 and R=1.1
\displaystyle \dfrac{1000}{500}\displaystyle =\displaystyle \dfrac{500 \times 1.1^{n-1}}{500}Divide both sides by 500
\displaystyle 2\displaystyle =\displaystyle 1.1^{n-1}Evaluate
\displaystyle \log 2\displaystyle =\displaystyle \log 1.1^{n-1}Take the logarithm of both sides
\displaystyle \log 2\displaystyle =\displaystyle (n-1) \log 1.1Use the \log power rule
\displaystyle \dfrac{\log 2}{\log 1.1}\displaystyle =\displaystyle \dfrac{(n-1) \log 1.1}{\log 1.1} Divide both sides by \log 1.1
\displaystyle 7.27\displaystyle =\displaystyle n-1 Evaluate with calculator
\displaystyle 7.27+1\displaystyle =\displaystyle n-1+1 Add 1 to both sides
\displaystyle 8.27\displaystyle =\displaystyle n Evaluate
\displaystyle n\displaystyle =\displaystyle 8.27 Make n the subject

If n=8.27, then Hayley will achieve more than \$1000 after the 8th week and exceed it at the beginning of 9th week.

Idea summary

We can apply the concept of geometric sequences to situations that involve exponential growth or exponential decay.

The initial condition

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 for the initial condition, referring to the first term of the sequence. However, sometimes it can be useful to use u_0 meaning the initial term. u_0 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 and each week this amount increases by \$20. If we want to know how much we have after 5 weeks and we use the initial condition of u_0=100 then 5 weeks later is u_5 where as if we used the initial condition of u_1=100 then 5 weeks later is u_6, which can be a bit confusing. Both would give the same answer but using u_0 in this case makes the term number we are looking for match the number of weeks.

Examples

Example 2

The average rate of depreciation of the value of a Ferrari is 14\% per year. A new Ferrari is bought for \$90 \, 000.

a

What is the car worth after 1 year?

Worked Solution
Create a strategy

Subtract the 1st year deduction from the initial price.

Apply the idea

For the 1st year deduction, multiply the initial price by the average rate 14 \% = 0.14.

\displaystyle \text{1st year deduction}\displaystyle =\displaystyle 90 \, 000 \times 0.14Multiply the initial price by the average rate
\displaystyle =\displaystyle 12 \, 600Evaluate
\displaystyle \text{Worth}\displaystyle =\displaystyle 90 \, 000 - 12 \,600Subtract 12 \,600 from the initial price
\displaystyle =\displaystyle \$ 77 \, 400Evaluate
b

What is the car worth after 3 years?

Worked Solution
Create a strategy

Multiply the initial price by the equivalent percentage of the remaining price.

Apply the idea

Getting the remaining price is the same as multiplying the initial price by its equivalent percentage.

So the remaning price is 100\%-14\%=86\%=0.86 of the initial price.

But finding the worth after 3 years is the same as raising 0.86 into 3, that is 0.86^3.

\displaystyle \text{Worth}\displaystyle =\displaystyle 90 \, 000 \times (0.86)^3Multiply initial price by 0.86^3
\displaystyle =\displaystyle 90 \, 000 \times 0.636056Evaluate the exponent
\displaystyle =\displaystyle \$ 57 \,245.04Evaluate
Reflect and check

Notice that 0.86 represents the worth of the car each year, this means it is the common ratio R.

So we can substitute R=0.86 and n=3+1=4 (since it is after 3 years) into u_n=aR^{n-1}.

\displaystyle u_n\displaystyle =\displaystyle aR^{n-1}Write the formula
\displaystyle u_4\displaystyle =\displaystyle 90 \, 000 \times 0.86^{4-1}Substitute n=4, a= 90\, 000, R= 0.86
\displaystyle =\displaystyle 90 \, 000 \times 0.86^3Evaluate the subtraction
\displaystyle =\displaystyle 90 \, 000 \times 0.636056Evaluate the exponent
\displaystyle =\displaystyle \$ 57 \,245.04Evaluate
c

Write a recursive rule for V_{n+1}, defining the value of the car after n years.

Write both parts of the rule on the same line, separated by a comma.

Worked Solution
Create a strategy

Substitite u=V, R=0.86 and a=90 \, 000 into u_{n+1}=Ru_n, u_0=a.

Apply the idea
\displaystyle u_{n+1}\displaystyle =\displaystyle Ru_n,u_0=aWrite the recursive formula
\displaystyle V_{n+1}\displaystyle =\displaystyle 0.86V_n, V_0=90\, 000Substitite u=V, R=0.86,a=90 \, 000

Example 3

The first blow of a hammer drives a post a distance of 64 \text{ cm} into the ground. Each successive blow drives the post \dfrac 34 as far as the preceding blow. In order for the post to become stable, it needs to be driven \dfrac{781}{4} \text{ cm} into the ground.

If n is the number of hammer strikes needed for the pole to become stable, find n.

Worked Solution
Create a strategy

Equate the required distance to the sum of the geometric series formula a\times \dfrac{1-r^n}{1-r}, where a=64, r=\dfrac34.

Apply the idea
\displaystyle \dfrac{781}{4}\displaystyle =\displaystyle a\times \dfrac{1-r^n}{1-r}Equate the required distance to a\times \dfrac{1-r^n}{1-r}
\displaystyle \dfrac{781}{4}\displaystyle =\displaystyle 64\times \dfrac{1-\left(\dfrac 34\right)^n}{1-\dfrac 34}Substitute a=64, r=\dfrac34
\displaystyle \dfrac{781}{4}\displaystyle =\displaystyle 64\times \dfrac{1-\left(\dfrac 34\right)^n}{\dfrac 14}Evaluate the denominator
\displaystyle \dfrac{781}{4} \times \dfrac 14\displaystyle =\displaystyle 64\times \dfrac{1-\left(\dfrac 34\right)^n}{\dfrac 14} \times \dfrac 14Multiply both sides by \times \dfrac 14
\displaystyle \dfrac{781}{16}\displaystyle =\displaystyle 64\times 1-\left(\dfrac 34\right)^nEvaluate
\displaystyle \dfrac{\dfrac{781}{16}}{64}\displaystyle =\displaystyle \dfrac{64\times 1-\left(\dfrac 34\right)^n}{64}Divide both sides by 64
\displaystyle \dfrac{781}{1024}\displaystyle =\displaystyle 1-\left(\dfrac 34\right)^nEvaluate
\displaystyle \dfrac{781}{1024}-1\displaystyle =\displaystyle 1-\left(\dfrac 34\right)^n-1Subtract 1 from both sides
\displaystyle \dfrac{781}{1024}-1\displaystyle =\displaystyle -\left(\dfrac 34\right)^nEvaluate
\displaystyle -1\times \left(\dfrac{781}{1024}-1\right)\displaystyle =\displaystyle -1\times \left(-\left(\dfrac 34\right)^n\right)Multiply both sides by -1
\displaystyle -\dfrac{781}{1024}+1\displaystyle =\displaystyle \left(\dfrac 34\right)^nEvaluate
\displaystyle 1-\dfrac{781}{1024}\displaystyle =\displaystyle \left(\dfrac 34\right)^nRearrange the left-hand side
\displaystyle \dfrac{243}{1024}\displaystyle =\displaystyle \left(\dfrac 34\right)^nEvaluate the subtraction
\displaystyle \dfrac{3^5}{4^5}\displaystyle =\displaystyle \left(\dfrac 34\right)^nWrite the fraction in exponential form
\displaystyle \left(\dfrac{3}{4}\right)^5\displaystyle =\displaystyle \left(\dfrac 34\right)^nApply the rule \dfrac{a^m}{b^m}=\left(\dfrac ab\right)^m
\displaystyle 5\displaystyle =\displaystyle nEquate exponents with the same base
\displaystyle n\displaystyle =\displaystyle 5Make n the subject
Idea summary

We can use the convention u_0 for the situations where we start with an initial value and then look for the amount in the days/weeks/month after the starting point.

Outcomes

U1.AoS2.5

the use of first-order linear recurrence relations of the form u_o=a,u_{n+1}=RU_nwhere 𝑎 and 𝑅 are constants to model compound interest investments and loans, reducing balance depreciation of an asset over time, including the rule for the future value of the asset after 𝑛 depreciation periods, compound interest investments and debts

U1.AoS2.7

use a given recurrence relation to generate a sequence, deduce the explicit rule, n u from the recursion relation, tabulate, graph and evaluate the sequence

U1.AoS2.9

demonstrate the use of a recurrence relation to determine the linear depreciating value of an asset after 𝑛 time periods for the initial sequence

U1.AoS2.10

use a rule for the future value of a linear depreciating asset to solve practical problems

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