Annuities are a number of equal payments made at regular intervals for a certain amount of time. There are two types of annuities: future value annuity and present value annuity.
In this post, we look at the future value annuity. Future value annuities are payments made into an investment or savings account resulting in an accumulated amount at the end of a time period.
We will derive the following formula and its modifications:
$${\normalsize F= \frac{x\left[(\,1+i)\,^{n}-1\right]}{i}}$$Grade 12 financial maths requires that you understand the derivation and the application of the different cases that you will see below.
We derive the modifications by using case studies:
- Payments made at the end of a term
- Payments made at the beginning of a term
- Payment made immediately and thereafter made at the end of each term.
- Payments made at the end of a term but missed the last payment.
Please note we will use timelines like we used them in grade 11 financial maths. If you have forgotten them, do not worry, the work here is very simple.
1. Payments made at the end of a term.
In this situation, one invests money at the end of a term(a month, a quarter, a semester or a year). In most cases, you will find that the instalments are made at the end of a month.
Let us now try a situation where one invests money, \(x\), at the end of a term, and the money grows at interest, \(i\), for a certain number of years. The timeline representing the situation is as follows:
The accumulated amount of this investment is:
$${\normalsize x(\,1+i)\,^2+x(\,1+i)\,+x(\, 1+i)\,^0 \text{ or }}$$ $${\normalsize x(\,1+i)\,^2+x(\,1+i)\,+x}$$Rearranging:
$${\normalsize x+x(\,1+i)\,+x(\,1+i)\,^2}$$This looks like a geometric series. We will then need to find the variables that are required for a geometric series, i.e., \(a\) and \(r\).
$$ \normalsize \begin{align*} a &= x\\ r &= \frac{x(\,1+i)\,}{x}\\ &=1+i, r>1 \end{align*} $$The sum of the geometric series formula that we will use is:
$${\normalsize S_n= \frac{a(\,r^{n}-1)\,}{r-1}}$$Now, substituting the variables obtained earlier into the above.
$$ \normalsize \begin{alignat*}{1} S_n &= \frac{x\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ S_n &= \frac{x\left[(1+i)^{n}-1\right]}{i} \end{alignat*} $$When we define the sum, \(\normalsize S_n\), as the future value, \(\normalsize F\), the above equation becomes:
$${\normalsize F= \frac{x\left[(\,1+i)\,^{n}-1\right]}{i}}$$where \(\normalsize F\) is the future value, \(\normalsize x\) is the monthly installment, \(\normalsize i\) is the interest, and \(\normalsize n\) is the number of instalments.
Example: Let us see if the two methods used above can give us the same answer if we invest R1000 at the end of each year. How much money will we have at the end of three years if the bank gives us interest of 10% p.a?
The first method requires us to to note that \(\normalsize i=10\% = 0.1\) and \(\normalsize x=R1000\). The answer is:
$$ \normalsize \begin{align} & x+x(\,1+i)\,+x(\,1+i)\,^2\\ &=1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ &=R3310 \end{align} $$Before using the formula, let us give you a tip on the \(\normalsize n\), the number of installements.
The \(\normalsize n\) is normally calculated as the number of years multiplied by the number of times the interest is calculated (1 for per year, 12 for months, 4 for per quarter, etc.)$${\normalsize n =last-first+1}$$
The number of instalments are not always easy to find, especially for a large sample. So when we have drawn the appropriate timeline, we will be able to get them right. For our example:
$${\normalsize n =3-1+1}$$Now
$$ \normalsize \begin{align} F &= \frac{x\left[(\,1+i)\,^{n}-1\right]}{i}\\ &= \frac{1000\left[(\,1+0.1)\,^{3}-1\right]}{0.1}\\ &=R3310 \end{align} $$Now let's summarise the information here. When payments are made at the end of each term (end of a year, month, quarter, six months, etc.), we use this equation this equation as is: $${\normalsize F=\frac{x[(x-1)^n-1]}{i}}$$
Remember
\(\normalsize F\) is the future value
\(\normalsize x\) is the installment
\(\normalsize i\) is the interest rate
\(\normalsize n\) is the number of instalments
2. Payments made at the beginning of a term.
Using the parameters of the first case, the timeline for the second case is as follows
The accumulated amount of this investment is:
$${\normalsize x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)\,}$$Rearranging in ascending order
$${\normalsize x(1+i)+x(1+i)^2+x(1+i)^3}$$Now, in this series:
$$ \normalsize \begin{align*} a&=x(\,1+i)\,\\ r&=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ &=1+i, r>1 \end{align*} $$Substituting into the sum of geometric series formula and using F instead of \(\normalsize S_n\):
$$ \normalsize \begin{align} S_n &= \frac{a(\,r^{n}-1)\,}{r-1}\\ F &= \frac{x(1+i)\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F &=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i) \end{align} $$Now let us check if we get the same solutions. For manual calculations:
$$ \normalsize \begin{align} F&=x(\,1+i)\,+x(\,1+i)\,^2+x(\,1+i)\,^3\\ &=1000(\,1+0.1)\,+1000(\,1+0.1)\,^2+1000(\,1+0.1)\,^3\\ &=R3641 \end{align} $$Using the formula:
$${n=2-0+1=3}$$The last payment was made at the beginning of year 3 which is at \(\normalsize T_2\). The first payment was made at the beginning of year 1, which is \(\normalsize T_0\). Hence we got three instalments which correspond to the timeline.
$$ \normalsize \begin{align} F&=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ &=\frac{1000\left[(1+0.1)^{3}-1\right]}{0.1}(1+0.1)\\ &=R3641 \end{align} $$Let's explain formula:$${\normalsize F=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\}$$
This formula is the one we got in case 1 but multiplied by \(\normalsize (1+i)\). This accounts for the last instalments being missed.
3. Payment made immediately and thereafter made at the end of each term
The timeline for this situation is displayed below:
The accumulated amount of this investment is:
$$ \normalsize \begin{align} &x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)\,^1+x(\, 1+i)^0\,\\ &=x(\,1+i)\,^3+x(\,1+i)^2\,+x(\, 1+i)^1+x \end{align} $$For the geometric series formula, and having rearranged the above:
$$ \normalsize \begin{align*} a&=x\\ r&=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ &=1+i, r>1 \end{align*} $$The sum of geometric series formula
$$ \begin{align} S_n &= \frac{a(\,r^{n}-1)\,}{r-1}\\ F &= \frac{x\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F &=\frac{x\left[(1+i)^{n}-1\right]}{i} \end{align} $$Reminder
\(\normalsize n\) is the number of instalments.
Checking the solutions, the manual calculation gives:
$$ \normalsize \begin{align} F &=x+x(\,1+i)\,+x(\,1+i)\,^2+x(\, 1+i)^3\\ &=1000+1000(\,1+0.1)\,+1000(\,1+0.1)\,^2+1000(\,1+0.1)\,^3\\ &=R4641 \end{align} $$Using the formula:
$${\normalsize n=3-0+1=4}$$Indeed there are four instalments in this situation.
$$ \normalsize \begin{align} F &= \frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ &= \frac{1000\left[(1+0.1)^{4}-1\right]}{0.1}(1+0.1)\\ &= R4641 \end{align} $$NB: Others prefer using the following equation when payment is made immediately:
$${\normalsize F=\frac{x\left[(1+i)^{n+1}-1\right]}{i}}$$The \(\normalsize n\) here is calculated as the normal number of years multiplied by the rate at which the interest is quoted.
Payments made at the end of a term but missed the last payment.
The timeline for this situation is displayed below:
We touched on a situation like in case 2 above. The accumulated amount of this investment is:
$${\normalsize x(\,1+i)\,^2+x(\,1+i)\,^1}$$Rearranging:
$${x(\, 1+i)\,+x(\,1+i)\,^2}$$For the sum of the geometric series formula::
$$ \normalsize \begin{align*} a &=x(\, 1+i)\,\\ r &=\frac{x(\,1+i)\,^2}{x(\,1+i)\,}\\ &=1+i, r>1 \end{align*} $$Then
$$ \normalsize \begin{align} S_n &= \frac{a(\,r^{n}-1)\,}{r-1}\\ F &= \frac{x(\, 1+i)\,\left[(1+i)^{n}-1\right]}{(1+i)-1}\\ F &=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i) \end{align} $$Now let us check the solutions. The manual calculation:
$$ \normalsize \begin{align} &x(\, 1+i)\,+x(\,1+i)\,^2\\ &=1000(\,1+0.1)\,+1000(\,1+0.1)\,^2\\ &=R2310 \end{align} $$Using the formula:
$${\normalsize n=2-1+1=2}$$ $$ \normalsize \begin{align} F&=\frac{x\left[(1+i)^{n}-1\right]}{i}(1+i)\\ &=\frac{1000\left[(1+0.1)^{2}-1\right]}{0.1}(1+0.1)\\ &=R2310 \end{align} $$This question shows why you shouldn't ignore the timeline when solving grade 12 financial math. They will help you get the correct number of instalments.