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Math help from the Learning Centre

This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

General Annuities

General annuities are essentially the opposite of simple annuities. Recall, for simple annuities, the frequency of the payment periods and compounding periods are the same. The defining characteristic of general annuities is that the frequency of the payment periods and compounding periods ARE NOT the same. Note that in this section, we are still only looking at simple annuities. Thus, there is still no payment at the beginning of the annuity. 

The steps to solve general annuity problems use knowledge from previous lessons. The main idea is to turn a general annuity problem into a simple annuity problem. There are 2 steps to solving general annuity problems: 

  1. Find the equivalent rate of interest for the payment frequency
  2. Use this interest rate in the correct simple annuity formula

For a refresher on simple annuities, click here


Example 1

Jamie deposits $250 into an account at the end of each month into an account that makes 5.75% compounded quarterly. How much money will be in the account after 5 years?



This is a simple general annuity because the deposits are made at the end of each month (simple) and the frequency of the payments, monthly, is not the same as the frequency of the compounding periods, quarterly (general). We are asked to find the future value of the annuity

First, we find the interest rate compounded monthly which is equivalent to 5.75% compounded quarterly. 

\begin{align} \left(1+\frac{r}{12}\right)^{12} &= \left(1+\frac{0.0575}{4}\right)^4 \\ \left[\left(1+\frac{r}{12}\right)^{12}\right]^{\frac{1}{12}} &= \left[\left(1+\frac{0.0575}{4}\right)^4\right]^{\frac{1}{12}} \\ 1+\frac{r}{12} &= \left(1+\frac{0.0575}{4}\right)^{\frac{4}{12}} \\ \frac{r}{12} &= \left(1+\frac{0.0575}{4}\right)^{\frac{4}{12}} -1 \\ r &= 12\left[\left(1+\frac{0.0575}{4}\right)^{\frac{4}{12}} -1\right] \\ r &= 0.057226659 \end{align}

Thus, the equivalent interest rate is 5.7226659%. Now, we can use this in the simple annuity formula. 

\(S_n = ? | R=$250 | i=\frac{5.7226659%}{12} | n=5*12=60\)

\(S_n = \frac{250[(1+\frac{0.057226659}{12})^{60}-1]}{\frac{0.057226659}{12}} = $17,318.75$\)

Thus, the account will have $17,318.75 after 5 years

Example 2

A car was purchased with monthly payments of $325 for 7 years. If interest is 8.5% compounded semi-annually, what is the purchase price of the car?



This is a general annuity question because the frequency of payments (monthly) is different than the frequency of compounding periods (semi-annually). First, calculate the interest rate compounded monthly which is equivalent to 8.5% compounded semi-annually. Here, we are looking for the present value. 

\begin{align} \left(1+\frac{r}{12}\right)^{12} &= \left(1+\frac{0.085}{2}\right)^2 \\ \left[\left(1+\frac{r}{12}\right)^{12}\right]^{\frac{1}{12}} &= \left[\left(1+\frac{0.085}{2}\right)^2\right]^{\frac{1}{12}} \\ 1+\frac{r}{12} &= \left(1+\frac{0.085}{2}\right)^{\frac{2}{12}} \\ \frac{r}{12} &= \left(1+\frac{0.085}{2}\right)^{\frac{2}{12}} -1 \\ r &= 12\left[\left(1+\frac{0.085}{2}\right)^{\frac{2}{12}} -1\right] \\ r &= 0.0835532745 \end{align}

Thus, the equivalent interest rate is 8.3532745%

\(A_n = ? | R=$325 | i = \frac{8.3532745%}{12} | n=7*12=84\)

\(A_n = \frac{325[(1-(1+\frac{0.0835532745}{12})^{-84}]}{\frac{0.0835532745}{12}} = $20,618.16\)

Thus, the purchase price of the car was $20,618.16


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