Using an Explicit Rule for Geometric Growth or Decay Revision Notes for VCE SSCE General Mathematics

Using an Explicit Rule for Geometric Growth or Decay

Introduction

When working with geometric sequences, you can calculate any term directly without needing to find all the previous terms first. This is done using an explicit rule, which provides a formula to jump straight to the value you need. This approach is particularly useful for solving real-world problems involving compound interest, investments, and asset depreciation.

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The advantage of using an explicit rule is that you can find any term in the sequence immediately, without having to calculate all the terms that come before it. This saves significant time, especially when working with large values of $n$ .

Deriving the general explicit rule

Consider an investment of $2000 that earns 5% compound interest per annum. Using a recurrence relation, you can model this as:

$V_0 = 2000, \quad V_{n+1} = 1.05V_n$

By calculating the first few terms, you'll notice a pattern:

$V_0 = 2000$
$V_1 = 1.05V_0$
$V_2 = 1.05V_1 = 1.05(1.05V_0) = 1.05^2V_0$
$V_3 = 1.05V_2 = 1.05(1.05^2V_0) = 1.05^3V_0$
$V_4 = 1.05V_3 = 1.05(1.05^3V_0) = 1.05^4V_0$

Following this pattern, after $n$ years the value becomes:

$V_n = 1.05^n V_0$

Using this formula, you can find the investment value after any number of years. For example, after 18 years:

$V_{18} = 1.05^{18} \times 2000 = \$4813.24 \text{ (to the nearest cent)}$

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Notice how the exponent matches the number of time periods. After 2 years, we multiply by $1.05^2$ ; after 3 years, by $1.05^3$ , and so on. This pattern is the foundation of the explicit rule.

The general formula for geometric sequences

For any geometric growth or decay situation modelled by:

$V_0 = \text{starting value}, \quad V_{n+1} = RV_n$

The value after $n$ iterations is:

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$V_n = R^n \times V_0$

where:

$V_0$ is the initial value
$R$ is the common ratio (or growth/decay multiplier)
$n$ is the number of time periods
$V_n$ is the value after $n$ periods

Critical distinction:

When $R > 1$ , the sequence shows growth (values increase)
When $R < 1$ , the sequence shows decay (values decrease)

Writing explicit rules for geometric sequences

To write an explicit rule, you need to identify the starting value $V_0$ and the multiplier $R$ from the recurrence relation.

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Worked Example: Writing Explicit Rules

Write an explicit rule for $V_n$ in terms of $n$ for each of the following, then find $V_6$ :

a) $V_0 = 5, \quad V_{n+1} = 4V_n$

b) $V_0 = 10, \quad V_{n+1} = 0.5V_n$

Solution:

Part a (Growth):

Identify the values: $V_0 = 5$ and $R = 4$

Write the explicit rule:

$V_n = 4^n \times 5$

Substitute $n = 6$ :

$V_6 = 4^6 \times 5 = 20480$

Part b (Decay):

Identify the values: $V_0 = 10$ and $R = 0.5$

Write the explicit rule:

$V_n = 0.5^n \times 10$

Substitute $n = 6$ :

$V_6 = 0.5^6 \times 10 = 0.15625$

Notice that part (a) demonstrates growth since $R = 4 > 1$ , whilst part (b) demonstrates decay since $R = 0.5 < 1$ .

Compound interest applications

When money is invested or borrowed with compound interest, the value grows over time. The growth multiplier $R$ can be determined from the annual interest rate $r\%$ using:

$R = 1 + \frac{r}{100}$

This leads to the explicit formula for compound interest:

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Compound Interest Formula:

$V_n = \left(1 + \frac{r}{100}\right)^n \times V_0$

where:

$V_0$ is the principal (initial amount invested or borrowed)
$r$ is the annual interest rate as a percentage
$n$ is the number of years
$V_n$ is the value after $n$ years

Remember: The "+1" in the formula represents keeping the original principal, while $\frac{r}{100}$ represents the interest earned.

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Worked Example: Investment Calculations

Consider an investment described by the rule:

$V_n = 1.09^n \times 10\,000$

Let's answer several questions about this investment.

a) How much was initially invested?

The initial investment is the value of $V_0$ , which is $10,000.

b) What is the annual interest rate?

Since $R = 1.09 = 1 + \frac{r}{100}$

We have $\frac{r}{100} = 0.09$

Therefore $r = 9$

The annual interest rate is 9%.

c) Find the value after 4 years (to the nearest cent)

Substitute $n = 4$ into the rule:

$V_4 = 1.09^4 \times 10\,000$

$V_4 = 14\,115.816...$

The value after 4 years is $14,115.82 (to the nearest cent).

d) Find the total interest earned over 4 years

To find total interest, subtract the principal from the final value:

Interest $= `$ 14,115.82 - $`10\,000 = `$ 4,115.82$`

The total interest earned is $4,115.82.

e) Find the interest earned in the fourth year only

First calculate $V_3$ :

$V_3 = 1.09^3 \times 10\,000 = 12\,950.29 \text{ (to the nearest cent)}$

The interest earned in the fourth year is:

$V_4 - V_3 = 14\,115.82 - 12\,950.29 = 1\,165.53$

Interest earned in the fourth year is $1,165.53.

Alternative method: Calculate 9% of $V_3$ : $0.09 \times 12\,950.29 = 1\,165.53$

f) Has the investor doubled their money within 10 years?

Calculate $V_{10}$ :

$V_{10} = 1.09^{10} \times 10\,000 = 23\,673.64$

Double the principal would be $2 \times 10\,000 = 20\,000$

Since $23,673.64 > $20,000, yes, the investor has doubled their money within 10 years.

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Exam Tip: When finding interest earned in a specific year, remember that you need to find the difference between consecutive terms, not just calculate interest on the original principal. This is because compound interest means you earn interest on previously earned interest.

Reducing balance depreciation

When an asset loses value over time, we model this using reducing balance depreciation. The decay multiplier $R$ is found using the depreciation rate $r\%$ :

$R = 1 - \frac{r}{100}$

This gives the explicit formula for reducing balance depreciation:

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Reducing Balance Depreciation Formula:

$V_n = \left(1 - \frac{r}{100}\right)^n \times V_0$

where:

$V_0$ is the purchase price of the asset
$r$ is the annual depreciation rate as a percentage
$n$ is the number of years
$V_n$ is the value after $n$ years

Note that for depreciation, $R < 1$ because the value is decreasing. The "-1" in the formula represents keeping part of the original value, while $\frac{r}{100}$ represents the proportion lost to depreciation.

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Worked Example: Machine Depreciation

A machine costs $9,500 and depreciates at 20% per year using the reducing balance method. The recurrence relation is:

$V_0 = 9500, \quad V_{n+1} = 0.8 \times V_n$

a) Write the explicit rule for the machine's value after $n$ years

First, identify the values:

$V_0 = 9500$

$R = 1 - \frac{20}{100} = 0.8$

The explicit rule is:

$V_n = 0.8^n \times 9500$

b) Find the machine's value after 8 years (to the nearest cent)

Substitute $n = 8$ :

$V_8 = 0.8^8 \times 9500$

$V_8 = 1593.835...$

After 8 years, the machine is worth $1,593.84 (to the nearest cent).

c) Calculate the total depreciation after 8 years

Total depreciation is the difference between the original value and the current value:

Depreciation $= `$ 9,500 - $`1,593.84 = `$ 7,906.16$`

The machine has depreciated by $7,906.16 after 8 years.

Finding unknown values using a calculator

You can use the explicit rule formulas to find unknown variables by solving equations. Whilst trial and error works, a CAS (Computer Algebra System) calculator provides exact solutions efficiently.

Finding the time period ( $n$ )

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Worked Example: Finding Time Period

How many years will it take for an investment of $5,000 at 6% per annum compound interest to grow above $8,000?

Solution:

Identify the known values:

$V_0 = 5000$

$V_n = 8000$

$R = 1 + \frac{6}{100} = 1.06$

Substitute into the explicit rule:

$V_n = R^n \times V_0$

$8000 = 1.06^n \times 5000$

Solve for $n$ using a calculator:

solve $(8000 = 1.06^n \cdot 5000, n)$

$n = 8.0661...$

After 8 years, the investment is worth $7,969.24.

Since we need the value to grow above $8,000, and interest is paid at the end of each year, the investment will exceed $8,000 after 9 years.

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Exam Tip: When asked how long until a value exceeds a certain amount, remember to round up to the next whole year since interest is typically added at the end of each year. The value of $n = 8.0661...$ means the investment reaches $8,000 partway through year 9.

Finding the interest or depreciation rate ( $r$ )

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Worked Example: Finding Depreciation Rate

A weaving company purchased a loom for $56,000. After 10 years, it has an estimated value of $15,000. If reducing balance depreciation is used, what is the annual depreciation rate? (Give your answer to one decimal place.)

Solution:

Identify the known values:

$V_0 = 56\,000$

$V_n = 15\,000$

$n = 10$

$R = 1 - \frac{r}{100}$

Substitute into the explicit rule:

$V_n = R^n \times V_0$

$15\,000 = \left(1 - \frac{r}{100}\right)^{10} \times 56\,000$

Solve for $r$ using a calculator:

solve $\left(15000 = \left(1 - \frac{r}{100}\right)^{10} \cdot 56000, r\right)$

The calculator gives two solutions:

$r = 12.342... \text{ or } r = 187.657...$

Choose the more appropriate answer: $r = 12.3\%$ (to one decimal place)

Note: The second solution (187.657...) is mathematically valid but not practical for depreciation, which is why we select the first answer.

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Exam Tip: When solving for percentage rates, calculators may give multiple solutions. Always choose the value that makes practical sense in the context of the problem. For depreciation rates, values should typically be between 0% and 100%.

Remember!

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Key Points to Remember:

The explicit rule for geometric sequences is $V_n = R^n \times V_0$ , where $V_0$ is the starting value and $R$ is the common ratio
For growth, $R > 1$ ; for decay, $R < 1$
Compound interest formula: $V_n = \left(1 + \frac{r}{100}\right)^n \times V_0$ where $r$ is the percentage interest rate
Reducing balance depreciation formula: $V_n = \left(1 - \frac{r}{100}\right)^n \times V_0$ where $r$ is the percentage depreciation rate
You can use the explicit rule to find any unknown variable (value, time, or rate) by substituting known values and solving the resulting equation
When finding interest or depreciation for a specific year, calculate the difference between consecutive terms, not the percentage of the original amount

Using an Explicit Rule for Geometric Growth or Decay (VCE SSCE General Mathematics): Revision Notes