If $300 is invested at an interest rate of 4.5% per year, find the amount of the investment at the end of 11 years for the following compounding methods.

Question

(a) Annually
$

(b) Semiannually
$

(c) Quarterly
$

(d) Continuously
$

Answer 1

A=P(1+r/n)^nt for compound interest.

a) A = 300 (1+0.045/1)^1*11=486.86

b) A = 300 (1+0.045/2)^2*11=489.46

c) A = 300 (1+0.045/4)^4*11=490.79

A = Pe^rt for continous interest.

d) A = 300e^0.045*11=492.15