An ordinary annuity is a series of regular payments made at the end of the period.

The future value of ordinary is the future value of a stream of equal payments made at the end of the period.

It is calculated as follow:

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]$$

Where,

C is the period of cash flow or payments.

i is the annual interest rate expressed in percentage

m is the number of times compounding occurs yearly

n is the number of years

To better understand this formula, we take three examples

__Example 1__

Supposed that N1500 is deposited at the end of each year for the next 6 years in an account paying 8% per year compounded annually. How much will be in the account at the end of the 6th year?

**Solution:**

Remember that,

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]$$

Here C=1500, m=1 because compounding occurs annually, i=0.08, n=6

$FV=1500\left[\frac{\left(1+\frac{0.08}{1}\right)^{6(1)}-1}{\frac{0.08}{1}}\right]$

$FV=1500\left[\frac{\left(1+0.08\right)^{6}-1}{0.08}\right]$

$FV=1500\left[\frac{\left(1.08\right)^{6}-1}{0.08}\right]$

$FV=1500\left[\frac{1.586874-1}{0.08}\right]$

$FV=1500\left[\frac{0.586874-1}{0.08}\right]=N11003.89$

__Example 2__

Find the future value of a 3-year annuity if the periodic payment is 1000 and the annuity earns 6% interest compounded semi-annually

**Solution:**

Recall that

$$FV=C\left[\frac{\left(1+\frac{i}{m}\right)^{n m}-1}{\frac{i}{m}}\right]$$

$FV=1000\left[\frac{\left(1+\frac{0.06}{2}\right)^{3(2)}-1}{\frac{0.06}{2}}\right]$

** Note: **m is 2 because there are only 2 6-month in a year.

$FV=1000\left[\frac{\left(1.03\right)^{6}-1}{0.03}\right]$

$FV=1000\left[\frac{\left(1.19405\right)-1}{0.03}\right]$

$FV=1000\left[\frac{0.19405}{0.03}\right]=N6568$

__Example 3__

Kenny wishes to determine how much depositing N1500 at the end of the period will amount to in the next 3 years. If the interest rate is 12% compounded quarterly, how much will be in his account?

**Solution:**

$FV=1500\left[\frac{\left(1+\frac{0.12}{4}\right)^{3(4))}-1}{\frac{0.12}{4}}\right]$

** Note**: m is 4 because a year is equivalent to 4 quarters.

$FV=1500\left[\frac{\left(1.03\right)^{12}-1}{0.03}\right]$

$FV=1000\left[\frac{\left(1.42576\right)-1}{0.03}\right]$

$FV=1000\left[\frac{0.042576}{0.03}\right]=N21,288$

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