Compound Interest - SS3 Mathematics Lesson Note
Most banks and corporate bodies utilize compound interest to compute important financial transactions such as loans and investment. Simple interest is calculated on the principal, or original, amount of a loan. Compound interest is calculated on the principal amount and the accumulated interest of previous periods, and thus can be regarded as “interest on interest.”
Using the Amount formula,
\[A = P{(1 + R)}^{T}\]
\(A\) = Amount of Future Value
\(P\) = Principal
\(R\) = Annual Rate
\(T\) = Number of years
\[Compound\ interest\ = \ Amount - Principal\]
Example 1 Find the future value on \(N6,000\ \)at \(10\%\) per annum for \(5\) years compounded annually.
Solution
\[A = P(1 + R)^{T}\]
\[P = 6000\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 10\% = 0.1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = 5\]
\[A = 6000{(1 + 0.1)}^{5}\]
\[A = 6000{(1.1)}^{5}\]
\[A = N9,663\]
Peradventure, the compound interest is meant to be calculated more frequently than once a year, usually semiannually, quarterly etc. the value \(n\) (number of times per annum interest is to be compounded) and the following table is considered:
|
TERM |
|---|
VALUE OF \(\mathbf{n}\)
Semiannually
\[2\]
Quarterly
\[4\]
Monthly
\[12\]
Daily
\[365\]
Then, \(A = P\left( 1 + \frac{R}{n} \right)^{nT}\)
Example 2 Find the future value on \(N3,000\) at \(10\%\) per annum for \(3\) years compounded semiannually.
Solution
\[A = P\left( 1 + \frac{R}{n} \right)^{nT}\]
\[P = 3000\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = 10\% = 0.1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n = 2\ \ \ \ \ \ \ \ \ \ \ \ \ T = 3\]
\[A = 3000\left( 1 + \frac{0.1}{2} \right)^{2(3)}\]
\[A = 3000(1 + 0.05)^{6}\]
\[A = 3000(1.05)^{6}\]
\[A = N4,020\]