5.05 Applications of geometric series: annuities

An annuity is a compound interest investment to which regular payments are made or from which regular payments are received for a fixed period of time. There are two types of annuities to consider: an investment account with regular, equal contributions and interest compounding at the end of each period, and a single-sum investment from which regular, equal withdrawals are made.

Making regular contributions

Here is an example where regular, equal deposits are made into an investment account and interest is compounded at the end of each period.

At the beginning of each year for $3$3 years, $\$500$$500 is deposited into an investment account that earns $3%$3% per annum interest.

• The first $\$500$$500 accrues interest for $3$3 years.
• The second $\$500$$500 accrues interest for $2$2 years.
• The third $\$500$$500 accrues interest for $1$1 year.

 

Notice that the funds in the account will increase quickly because of the regular contributions and the interest paid on the balance. We are interested in the total value of the investment after $3$3 years.

Since each equal, regular contribution earns compound interest, we can use the formula on each contribution separately to find the future value of the annuity as a whole.

From the example above:

• the first contribution will have a final value of $\$500\times1.03^3$$500×1.033
• the second will have a final value of $\$500\times1.03^2$$500×1.032
• the last contribution will have a final value of $\$500\times1.03$$500×1.03 by the end of the investment

However, for longer term investments, calculating the total in this way would be tedious. Take a close look at the series created by the investment $\$500\times1.03$$500×1.03 +$\$500\times1.03^2$$500×1.032 + $\$500\times1.03^3$$500×1.033. This is a geometric series. So, we can use the formula for the sum of a geometric sequence for these problems.

Worked example

Example 1

A bank client deposits $\$1000$$1000 at the beginning of each year, and is given $7%$7% interest per year for $50$50 years. How much will accrue in the account over that time?

Think: To develop a geometric sequence, let's consider the money that is in the account at the end of each year. Rather than calculating each amount, let's leave them in expanded form as this will allow us to see any patterns. If we set $A_n$An​ as the amount of money at the end of the $n$nth year, then we have:

$A_0=1000$A0​=1000

And after one year, since $7%$7% interest has been added, we have:

$A_1$A1​

$=$=

$1000\times\left(1.07\right)$1000×(1.07)

By the end of the second year, another $\$1000$$1000 has been added and receives interest, but the original $\$1000$$1000 has been boosted by another interest payment too.

The total amount can be determined by:

$A_2$A2​	$=$=	$(A_1+1000)(1.07)$(A1​+1000)(1.07)
$A_2$A2​	$=$=	$1000\left(1.07\right)\left(1.07\right)+1000\left(1.07\right)$1000(1.07)(1.07)+1000(1.07)
$A_2$A2​	$=$=	$1000\left(1.07^2+1.07\right)$1000(1.072+1.07)

By the end of the third year, the total accrual becomes:

$A_3=1000\left(1.07^3+1.07^2+1.07\right)$A3​=1000(1.073+1.072+1.07)

Do: A pattern is emerging, and by the end of $50$50 years, the total accrued becomes:

$A_{50}=1000\left(1.07^{50}+1.07^{49}+1.07^{48}+...+1.07\right)$A50​=1000(1.0750+1.0749+1.0748+...+1.07)

It might help to reverse the terms inside the brackets, so that we can see that we have is a geometric series with first term and common ratio both equal to $1.07$1.07:

$A_{50}=1000\left(1.07+1.07^2+1.07^3+...+1.07^{50}\right)$A50​=1000(1.07+1.072+1.073+...+1.0750)

Hence, we can now state $a=1000$a=1000, $r=1.07$r=1.07, $n=50$n=50 and substitute into the formula $A_n=1000\times S_n$An​=1000×Sn​, where $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn​=a(rn−1)r−1​. So we have:

$A_n$An​	$=$=	$1000\left(\frac{a\left(r^n-1\right)}{r-1}\right)$1000(a(rn−1)r−1​)
$A_n$An​	$=$=	$1000\left(\frac{1.07\left(1.07^{50}-1\right)}{1.07-1}\right)$1000(1.07(1.0750−1)1.07−1​)
$A_n$An​	$\approx$≈	$\$434985.95$$434985.95

Hence the amount accrued in the account will be approximately $\$434985.95$$434985.95.

We can use a different method to model the annuity, finding the final value of each investment at the end of the time period and adding them together. This will result in the same series as you find by calculating the value in the single investment account at the end of the total time period. Let's use this method in the example below, where the unknown is not the amount earned, but the size of each investment.

Example 2

Emily wishes to save $\$2000$$2000 throughout Year $11$11 and $12$12 for her schoolies trip. Beginning in January of Year $11$11, she deposits a certain amount into her bank account at the start of each month. Interest is paid at the end of each month, at a rate of $4%$4% p.a. compounding monthly. If she wishes to withdraw the $\$2000$$2000 at the end of November in Year $12$12, what is the size of each monthly deposit?

Think: before we develop our sequence, we need to find the values of $n$n and $r$r.

If she makes deposits from January until November of the following year, this means $n=23$n=23. $4%$4% p.a. becomes $\frac{4}{12}%$412​% per month, so the $r$r for our geometric series will be:

$1+\frac{4}{1200}=1\frac{1}{300}$1+41200​=11300​

Do: In this example, we will let $A_n$An​ be the value of the $n$nth deposit at the end of November of Year $12$12. Let $M$M be the value of each monthly deposit.

Thus, $A_1=M\times(1\frac{1}{300})^{23}$A1​=M×(11300​)23, since the first deposit accrues interest over $23$23 months.

$A_2=M\times(1\frac{1}{300})^{22}$A2​=M×(11300​)22

$A_3=M\times(1\frac{1}{300})^{21}$A3​=M×(11300​)21

And so on, until we reach the last deposit:

$A_{23}=M\times(1\frac{1}{300})^1$A23​=M×(11300​)1

The sum of these $23$23 investments must equal $\$2000$$2000:

$2000=M\times(1\frac{1}{300})^1+M\times(1\frac{1}{300})^2+M\times(1\frac{1}{300})^3+\ldots+M\times(1\frac{1}{300})^{23}$2000=M×(11300​)1+M×(11300​)2+M×(11300​)3+…+M×(11300​)23

Factorising for $M$M, we can then add the geometric series inside the brackets with $a=1\frac{1}{300}$a=11300​, $r=1\frac{1}{300}$r=11300​ and $n=23$n=23:

$2000=M((1\frac{1}{300})^1+(1\frac{1}{300})^2+(1\frac{1}{300})^3+\ldots+(1\frac{1}{300})^{23})$2000=M((11300​)1+(11300​)2+(11300​)3+…+(11300​)23)

$2000=M\times1\frac{1}{300}(\frac{(1\frac{1}{300})^{23}-1}{1\frac{1}{300}-1})$2000=M×11300​((11300​)23−111300​−1​)

Rearranging to make $M$M the subject, we get:

$M=\$83.52$M=$83.52

Here are some tips and tricks that may help you in solving these problems:

• Remember, if the compounding period is not annual, you will need to divide the interest rate and multiply the number of time periods by the relevant value.
• The amount of interest earned is calculated by subtracting the sum of all contributed amounts from the final value of the investment.

Practice questions

Question 1

question 2

Question 3

Analysing our investment with a recurrence relation

Because of the pattern of regular and equal investments associated with an annuity, we can model them using a recurrence relation. To develop our first recursive model, let's look at the following scenario.

We'll invest $\$5000$$5000 at $6%$6% per annum interest compounded monthly. At the end of each month, after the interest has been applied, we'll make an additional deposit of $\$100$$100.

In order to develop a recursive model for this annuity, we first need to change our interest rate from annual to monthly by dividing by $12$12, so we get $0.5%$0.5% per month. At the end of each month, we increase the balance by $0.5%$0.5% ($\times1.005$×1.005) and then add $\$100$$100.

The balance at the end of the first few months may now be calculated as follows. (We are dealing with money, so we have rounded each balance to $2$2 decimal places and then use the rounded amount in the following row.)

End of month 1	$1.005\times5000+100$1.005×5000+100	$=$=	$\$5125$$5125
End of month 2	$1.005\times5125+100$1.005×5125+100	$=$=	$\$5250.63$$5250.63
End of month 3	$1.005\times5250.63+100$1.005×5250.63+100	$=$=	$\$5376.88$$5376.88

Notice a recurring pattern:

$\text{Balance at the end of the month }=\text{Balance at the beginning }\times1.005+100$Balance at the end of the month =Balance at the beginning ×1.005+100

The balance at the end of each month is carried forward to be the starting balance in the account. Hence, using the notation we previously encountered using recurrence relations, we can write this as:

$A_n=A_{n-1}\times1.005+100$An​=An−1​×1.005+100, $A_0=5000$A0​=5000.

To examine the balance in a few years time, we don't want to manually calculate the values in the table. We could use a spreadsheet, or use the formula for the sum of a geometric sequence as demonstrated previously.

Practice questions

Question 4

Question 5