In: Advanced Math
An investment account with an annual interest rate of 7% was opened with an initial deposit of $4,000 Compare the values of the account after 9 years when the interest is compounded annually, quarterly, monthly, and continuously.
Consider the Compound Interest formula;
A(t) = P(1 + r/n)nt
Where, A(t) is the account value, “t” is the measured in years, “P” is the starting amount or initial value, “r” is the annual percentage rate (APR), “n” is the number of compounding periods in one year.
Suppose the initial deposit is P = 4000 dollars.
The annual percentage rate is r = 7%, that is;
r = 7/100
= 0.07
The interest is compounded annually n = 1,
The investment amount after t = 9 years is determined as follows;
Substitute the known value as follows:
A(9) = 4000(1 + 0.07/1)1×9
= 4000(1.07)9
= 7353.84
Therefore, the amount after 9 years that compounded annually is 7353.84 dollars.
The interest is compounded quarterly n = 4,
The investment amount after t = 9 years is determined as follows;
Substitute the known value as follows:
A(9) = 4000(1 + 0.07/4)4×9
= 4000(1.0175)36
= 7469.63
Therefore, the amount after 9 years that compounded quarterly is 7469.63 dollars.
The interest is compounded monthly n = 12,
The investment amount after t = 9 years is determined as follows;
Substitute the known value as follows:
A(9) = 4000(1 + 0.07/12)12×9
= 4000(1.0058)108
= 7497
Therefore, the amount after 9 years that compounded monthly is 7497 dollars.
The formula for the continuous growth or decay is:
A(t) = aert
Here,
The initial deposit is a = 4000.
The continuous growth rate is r;
r = 7/100
= 0.07
The investment amount after t =9 years is determined as follows;
Substitute the known value as follows:
A(9) = 4000(e0.07×9)
= 4000(e0.63)
= 4000 × 1.8776
= 7510
Therefore, the investment account after 9 years is 7510 dollars.