Investments & Savings Revision Notes for Leaving Cert Mathematics

Investments & Savings

For savings account, deposits are put into the account at regular intervals, such as the beginning of every year, month, week etc. Each of these deposits will earn interest for a different period of time.

Example

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John puts €100 euro into his savings account every month for 5 years. How much money will be in the account after these 5 years.

First, calculate the monthly interest rate

(1+i)=(1.055)^\tfrac{1}{12}

Do not evaluate and round this expression yet, as you will lose precision in the calculation process.

The first $100$ euro is invested for $60$ months (5 years), then the second $100$ euro is invested for 59 months and so on up until the very last investment which stays in the account for 1 month. The final amount will be a sum of all these investments.

F=100(1.055^\tfrac{1}{12})^{60}+100(1.055^\tfrac{1}{12})^{59} +...+100(1.055^\tfrac{1}{12})^{1}

F=100(1.055^\tfrac{60}{12})+100(1.055^\tfrac{59}{12}) +...+100(1.055^\tfrac{1}{12})

Factor out $100$ .

F=100\left( (1.055^\tfrac{60}{12})+(1.055^\tfrac{59}{12}) +...+(1.055^\tfrac{1}{12}) \right)

Notice that within the large bracket, there is a geometric series. Rearrange the terms :

F=100\left( (1.055^\tfrac{1}{12})+(1.055^\tfrac{2}{12}) +...+(1.055^\tfrac{60}{12}) \right)

Now we can see the starting term $a$ , the common ration $r$ and the number of terms $n$ .

a=1.055^\tfrac{1}{12},r=1.055^{\tfrac{1}{12}},n=60

Plug these parameters into the sum of a geometric sequence.

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

S_{60}=\frac{1.055^\tfrac{1}{12}(1-(1.055^\tfrac{1}{12})^{60})}{1-1.055^\tfrac{1}{12}}

Remember that €100 we factored out as well :

100\left(S_{60}=\frac{1.055^\tfrac{1}{12}(1-(1.055^\tfrac{1}{12})^{60})}{1-1.055^\tfrac{1}{12}} \right) =:success[€6895.20]

Example

infoNote

What sum of money must you save each month in order to have €5500 in your savings account after 4 years at an annual rate of 4.25%?

Calculate the monthly interest rate

(1 + i) = (1.0425)^\tfrac{1}{12}

Do not evaluate and round this expression yet, as you will lose precision in the calculation process.

The monthly deposit, (P), will accumulate for 48 months (4 years). The first deposit will stay in the account for 48 months, the second deposit for 47 months, and so on, up to the last deposit, which remains in the account for 1 month. The total future value of the savings account is the sum of all these deposits:

5500 = P \cdot \left((1.0425^\tfrac{1}{12})^{48} + (1.0425^\tfrac{1}{12})^{47} + \dots + (1.0425^\tfrac{1}{12})^{1}\right)

Factor out (P):

5500 = P \cdot \left((1.0425^\tfrac{1}{12})^{1} + (1.0425^\tfrac{1}{12})^{2} + \dots + (1.0425^\tfrac{1}{12})^{48}\right)

Recognise the geometric series :

The terms inside the parentheses form a geometric series with:

First term: $a = (1.0425^\tfrac{1}{12})^{1}$ ,
Common ratio: $r = (1.0425^\tfrac{1}{12})$ ,
Number of terms: $n = 48$ . The sum of a geometric series is given by:

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

Substitute the values of $a,r,n$ .

S_{48} = \frac{(1.0425^\tfrac{1}{12})(1 - (1.0425^\tfrac{1}{12})^{48})}{1 - (1.0425^\tfrac{1}{12})}

The total amount, €5500 :

5500 = P \cdot \frac{(1.0425^\tfrac{1}{12})(1 - (1.0425^\tfrac{1}{12})^{48})}{1 - (1.0425^\tfrac{1}{12})}

P=:success[€109.67]

Investments & Savings (Leaving Cert Mathematics): Revision Notes