Investing Money by Regular Instalments (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example: Rawen's Birthday Investment

Question: Rawen's parents invested $1000 in his name on the day he was born. They continued investing $1000 on each birthday until his 20th birthday. On his 21st birthday they gave him the investment's value. If the money earned 7% interest compounded annually, what was the final value to the nearest dollar?

Solution:

First, identify what each instalment grows to:

The 1st instalment is invested for 21 years: $1000 \times 1.07^{21}$

The 2nd instalment is invested for 20 years: $1000 \times 1.07^{20}$

Continue this pattern...

The 20th instalment is invested for 2 years: $1000 \times 1.07^2$

The 21st and last instalment is invested for 1 year: $1000 \times 1.07^1$

The total amount $A_{21}$ after 21 years equals instalments plus interest:

$A_{21} = (1000 \times 1.07^1) + (1000 \times 1.07^2) + \cdots + (1000 \times 1.07^{21})$

This forms a GP with:

• First term: $a = 1000 \times 1.07$
• Common ratio: $r = 1.07$
• Number of terms: $n = 21$

Using the GP sum formula:

$A_{21} = \frac{a(r^{21} - 1)}{r - 1}$

$= \frac{1000 \times 1.07 \times (1.07^{21} - 1)}{0.07}$

$\approx \$48006$

Therefore, Rawen received $48,006 on his 21st birthday.

Worked Example: Robin and Robyn's Superannuation

Question: Robin and Robyn invest $10000 in a superannuation scheme on 1st July each year, beginning in 2010. The money earns 8% pa compound interest, compounded annually.

Part a: How much will the fund amount to by 30th June 2030?

Solution a:

The 1st instalment is invested for 20 years: $10000 \times 1.08^{20}$

The 2nd instalment is invested for 19 years: $10000 \times 1.08^{19}$

The 19th instalment is invested for 2 years: $10000 \times 1.08^2$

The 20th and last instalment is invested for 1 year: $10000 \times 1.08^1$

The total amount $A_{20}$ after 20 years:

$A_{20} = (10000 \times 1.08^1) + (10000 \times 1.08^2) + \cdots + (10000 \times 1.08^{20})$

This is a GP with first term $a = 10000 \times 1.08$, ratio $r = 1.08$, and 20 terms.

$A_{20} = \frac{a(r^{20} - 1)}{r - 1}$

$= \frac{10000 \times 1.08 \times (1.08^{20} - 1)}{0.08}$

$\approx \$494229$

Part b: How much will the fund amount to by the end of $n$ years?

Solution b:

The 1st instalment is invested for $n$ years: $10000 \times 1.08^n$

The 2nd instalment is invested for $n - 1$ years: $10000 \times 1.08^{n-1}$

The $n$th and last instalment is invested for 1 year: $10000 \times 1.08^1$

The total amount $A_n$ after $n$ years:

$A_n = (10000 \times 1.08^1) + (10000 \times 1.08^2) + \cdots + (10000 \times 1.08^n)$

This is a GP with first term $a = 10000 \times 1.08$, ratio $r = 1.08$, and $n$ terms.

$A_n = \frac{a(r^n - 1)}{r - 1}$

$= \frac{10000 \times 1.08 \times (1.08^n - 1)}{0.08}$

$= 135000 \times (1.08^n - 1)$

This general formula can be used for any number of years.

Part c: Show that 2031 is the year when the fund first exceeds $500000 on 30th June.

Solution c:

From part a, we know the total after 20 years is just under $500000.

Substituting $n = 21$ into the formula from part b:

$A_{21} = 135000 \times (1.08^{21} - 1)$

$\approx \$544568$

Therefore, 2031 is the year when the fund first exceeds $500000 on 30th June.

Part d: What annual instalment would have produced $1000000 by 2030?

Solution d:

Reworking part b with an instalment of $M instead of $10000:

$A_n = 13.5 \times M \times (1.08^n - 1)$

Substituting $n = 20$ and $A_{20} = 1000000$:

$1000000 = 13.5 \times M \times (1.08^{20} - 1)$

Making $M$ the subject:

$M = \frac{1000000}{13.5 \times (1.08^{20} - 1)}$

$\approx \$20234$

They would need to invest $20,234 annually to reach $1,000,000 by 2030.

Worked Example: Finding When the Fund Reaches $700,000

Using the previous example, find the year when the fund first exceeds $700000 on 30th June.

Solution:

Using the formula from part b with $M = 10000$ and $A_n = 700000$:

$700000 = 135000 \times (1.08^n - 1)$

Divide both sides by 135000:

$1.08^n - 1 = \frac{700}{135}$

Add 1 to both sides:

$1.08^n = \frac{835}{135}$

Convert to logarithmic form:

$n = \log_{1.08}\frac{835}{135}$

Use the change-of-base formula:

$n = \frac{\log_{10}\frac{835}{135}}{\log_{10}1.08}$

$\approx 23.68$

Since $n$ must be a whole number of years, the fund first exceeds $700,000 when $n = 24$, which is in 2034.

Worked Example: Charmaine's Superannuation Schemes

Question: Charmaine has a superannuation scheme with monthly instalments of $600 for 10 years and an interest rate of 7.8% pa, compounded monthly.

Part a: What will the final value be?

Solution a:

The monthly interest rate is: $0.078 \div 12 = 0.0065$

There are 120 months in 10 years.

The 1st instalment is invested for 120 months: $600 \times 1.0065^{120}$

The 2nd instalment is invested for 119 months: $600 \times 1.0065^{119}$

The 120th and last instalment is invested for 1 month: $600 \times 1.0065^1$

The total amount $A_{120}$ after 120 months:

$A_{120} = (600 \times 1.0065^1) + (600 \times 1.0065^2) + \cdots + (600 \times 1.0065^{120})$

This is a GP with first term $a = 600 \times 1.0065$, ratio $r = 1.0065$, and 120 terms.

$A_{120} = \frac{a(r^{120} - 1)}{r - 1}$

$= \frac{600 \times 1.0065 \times (1.0065^{120} - 1)}{0.0065}$

$\approx \$109257$

Part b: What weekly instalments would yield the same final value with 7.8% pa compounded weekly?

Solution b:

The weekly interest rate is: $0.078 \div 52 = 0.0015$

Let $M be the weekly instalment. There are 520 weeks in 10 years.

The 1st instalment is invested for 520 weeks: $M \times 1.0015^{520}$

The 2nd instalment is invested for 519 weeks: $M \times 1.0015^{519}$

The 520th and last instalment is invested for 1 week: $M \times 1.0015^1$

The total amount $A_{520}$ after 520 weeks:

$A_{520} = M \times 1.0015 + M \times 1.0015^2 + \cdots + M \times 1.0015^{520}$

This is a GP with first term $a = M \times 1.0015$, ratio $r = 1.0015$, and 520 terms.

$A_{520} = \frac{M \times 1.0015 \times (1.0015^{520} - 1)}{0.0015}$

$= \frac{M \times 10015 \times (1.0015^{520} - 1)}{15}$

Rearranging with $M$ as the subject:

$M = \frac{15 \times A_{520}}{10015 \times (1.0015^{520} - 1)}$

Since $A_{520}$ must equal the value from part a ($109257):

$M \approx \$138.65$

Part c: Which scheme costs more per year?

The weekly scheme costs approximately: $\$138.65 \times 52 \approx \$7210.04 \text{ per year}$

The monthly scheme costs: $\$600 \times 12 = \$7200 \text{ per year}$

Therefore, the weekly scheme costs slightly more per year (by about $10).