Finding Term n in a Sequence Modelling Geometric Growth and Decay (VCE SSCE General Mathematics): Revision Notes

Worked Example: Compound interest investment

A compound interest investment of $1000 earns interest at 10% per year. The recurrence relation is:

$V_0 = 1000, \quad V_{n+1} = 1.1V_n$

Find the value after 15 years, to the nearest dollar.

Solution:

We use the rule $V_n = R^n \times V_0$ where:

• $V_0 = 1000$
• $n = 15$
• $R = 1.1$

$V_n = R^n \times V_0$

$= 1.1^{15} \times 1000$

$= 4177.24...$

$= \$4177 \text{ (nearest dollar)}$

Result: The investment grows to $4177 after 15 years.

Worked Example: Reducing-balance depreciation

A car purchased for $18,500 depreciates at 10% per year. The recurrence relation is:

$V_0 = 18500, \quad V_{n+1} = 0.9V_n$

Find the value after 12 years, to the nearest dollar.

Solution:

We use the rule $V_n = R^n \times V_0$ where:

• $V_0 = 18500$
• $n = 12$
• $R = 0.90$

$V_n = R^n \times V_0$

$= 0.9^{12} \times 18500$

$= 5224.94...$

$= \$5225 \text{ (nearest dollar)}$

Result: The car's value decreases to $5225 after 12 years.

Worked Example: Comprehensive investment analysis

A principal of $20,000 is invested in an account earning compound interest at 6% per annum. The rule for the value after $n$ years is:

$V_n = 1.06^n \times 20000$

Part a: Find the value after 5 years, to the nearest cent.

Substitute $n = 5$ into the rule:

$V_5 = 1.06^5 \times 20000$

$= 26764.511552$

$= \$26764.51 \text{ (nearest cent)}$

Part b: Find the total interest earned after 5 years.

The interest earned equals the difference between the final value and the principal:

$\text{Interest} = V_5 - V_0$

$= 26764.51 - 20000$

$= \$6764.51$

Part c: Find the interest earned in the fifth year only.

The interest earned in the fifth year is the difference between the value at the end of year 5 and the value at the end of year 4.

First, calculate $V_4$:

$V_4 = 1.06^4 \times 20000 = 25249.54 \text{ (nearest cent)}$

Now find the difference:

$V_5 - V_4 = 26764.51 - 25249.54 = \$1514.97$

Part d: If interest compounds monthly instead of yearly, write a rule for the value after $n$ months.

When the compounding period changes, we need to adjust the interest rate accordingly:

Let $V_n$ be the value after $n$ months.

The monthly interest rate is: $r = \frac{6}{12} = 0.5\%$

The common ratio becomes: $R = 1 + \frac{0.5}{100} = 1.005$

Therefore, the rule is:

$V_n = 1.005^n \times 20000$

Part e: Find the value after 5 years (60 months).

Substitute $n = 60$:

$V_{60} = 1.005^{60} \times 20000 = \$26977.00$

Key observation: Notice that monthly compounding produces a higher final value than annual compounding ($26,977.00 vs $26,764.51). This is because interest is calculated and added more frequently.

Worked Example: Calculating credit card interest

Determine the interest payable on a credit card debt of $5630 at 17.8% per annum, compounding daily, for 27 days.

Solution:

First, calculate the total debt after 27 days using:

• $P = 5630$
• $r = 17.8$
• $n = 27$

$A = P\left(1 + \frac{r}{36500}\right)^n$

$= 5630\left(1 + \frac{17.8}{36500}\right)^{27}$

$= \$5704.60 \text{ (nearest cent)}$

Now calculate the interest:

$I = A - P$

$= 5704.60 - 5630.00$

$= \$74.60$

Result: The interest charged is $74.60 for just 27 days.

Worked Example: Credit card with interest-free period

Janelle pays for a holiday costing $1500 using her credit card. Her bank offers:

• 30-day statement period
• Additional 25 interest-free days
• 20% per annum interest after the interest-free period (compounding daily)

She makes the purchase on 17 August (day 10 of her statement period) and pays on 1 November. How much must she pay?

Solution:

Step 1: Calculate interest-free days

Days remaining in statement: $30 - 10 = 20$ days

Additional interest-free days: $25$ days

Total interest-free: $20 + 25 = 45$ days

Step 2: Count total days from purchase to payment

Starting from 18 August (day after purchase):

• August 18-31: 14 days
• September 1-30: 30 days
• October 1-31: 31 days
• Total: $14 + 30 + 31 = 75$ days

Step 3: Calculate days with interest

Interest payable days: $75 - 45 = 30$ days

Step 4: Calculate amount owed

Using $P = 1500$, $r = 20$, $n = 30$:

$A = 1500\left(1 + \frac{20}{36500}\right)^{30}$

$= \$1524.85 \text{ (nearest cent)}$

Result: Janelle must pay back $1524.85.

Worked Example: Comparing options for a $2000 purchase

Calculate the total cost under each option for a $2000 purchase:

Option a - Cash:

Pay $2000 at the time of purchase.

Option b - Simple-interest loan at 8% for 2 years:

Annual interest: $D = \frac{8}{100} \times 2000 =$160$

Total after 2 years: $V\_2 = 2000 + 160 \times 2 =$2320$

Option c - Compound-interest loan at 6% for 2 years:

Common ratio: $R = 1.06$

Total after 2 years:

$V_2 = 1.06^2 \times 2000 = \$2247.20$

Option d - Credit card (30-day statement + 15 interest-free days, 20% p.a., purchased day 15, paid after 2 years):

Interest-free days: $15 + 15 = 30$ days

Total days (2 years): $730$ days

Days with interest: $730 - 30 = 700$ days

$A = 2000\left(1 + \frac{20}{36500}\right)^{700} = \$2934.70$

Option e - Buy-now-pay-later ($30 initial fee + $10 monthly for 2 years):

Total fees: $30 + (24 \times 10) =$270$

Total cost: $270 + 2000 =$2270$

Option f - Best option when cash unavailable:

Comparing all options:

• Simple-interest loan: $2320
• Compound-interest loan: $2247.20 ✓ (cheapest)
• Credit card: $2934.70
• Buy-now-pay-later: $2270

Conclusion: The compound-interest loan is the most economical option at $2247.20.

Worked Example: Price changes over 2 years

A loaf of bread costs $2.20 at the end of 2021. Inflation is 2.7% in 2022 and 3.5% in 2023. Find the price at the end of 2023.

Solution:

Year 2022:

Price increase: $2.20 \times \frac{2.7}{100} = 0.06$

Price at end of 2022: $2.20 + 0.06 =$2.26$

Year 2023:

Price increase: $2.26 \times \frac{3.5}{100} = 0.08$

Price at end of 2023: $2.26 + 0.08 =$2.34$

Result: The bread price increases to $2.34 after 2 years.

Worked Example: Long-term price projection

A one-litre carton of milk costs $1.70 today. What will it cost in 20 years if average annual inflation is:

• Part a: 2.1%
• Part b: 6.8%

Solution:

This is equivalent to investing $1.70 at the given inflation rate, so we use:

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

Part a: With $P = 1.70$, $n = 20$, $r = 2.1$:

$\text{Price} = 1.70 \times \left(1 + \frac{2.1}{100}\right)^{20}$

$= \$2.58 \text{ (nearest cent)}$

Part b: With $P = 1.70$, $n = 20$, $r = 6.8$:

$\text{Price} = 1.70 \times \left(1 + \frac{6.8}{100}\right)^{20}$

$= \$6.34 \text{ (nearest cent)}$

Key observation: Notice the dramatic difference: at 2.1% inflation, the price increases by about 50%, but at 6.8% inflation, it nearly quadruples!

Worked Example: Calculating purchasing power

If $100,000 is stored for 8 years and average inflation is 3.7%, what is its purchasing power in current dollars?

Solution:

We need to solve:

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

for $P$, where $A = 100000$, $r = 3.7$, $n = 8$:

$100000 = P \times \left(1 + \frac{3.7}{100}\right)^8$

Using a calculator to solve for $P$:

$P = \frac{100000}{\left(1.037\right)^8} = \$74777 \text{ (nearest dollar)}$

Conclusion: This means $100,000 in 8 years' time has the same buying power as only $74,777 today if inflation averages 3.7% per year.