Answer
For $n=1$: $A=1828.49$
For $n=2$: $A=1830.29$
For $n=4$: $A=1831.19$
For $n=12$: $A=1831.80$
For $n=365$: $A=1832.09$
For continuous compounding: $A=1832.10$
Work Step by Step
$A=P(1+\frac{r}{n})^{nt}$ (n compounds per year)
$A=Pe^{rt}$ (continuous compounding)
$P=1500,~~n=1,~~t=10,~~r=0.02=2$%:
$A=1500(1+\frac{0.02}{1})^{1\times10}=1500(1.02)^{10}=1828.49$
$P=1500,~~n=2,~~t=10,~~r=0.02=2$%:
$A=1500(1+\frac{0.02}{2})^{2\times10}=1500(1.01)^{20}=1830.29$
$P=1500,~~n=4,~~t=10,~~r=0.02=2$%:
$A=1500(1+\frac{0.02}{4})^{4\times10}=1500(1.005)^{40}=1831.19$
$P=1500,~~n=12,~~t=10,~~r=0.02=2$%:
$A=1500(1+\frac{0.02}{12})^{12\times10}=1500(1+\frac{0.01}{6})^{120}=1831.80$
$P=1500,~~n=365,~~t=10,~~r=0.02=2$%:
$A=1500(1+\frac{0.02}{365})^{365\times10}=1500(1+\frac{0.02}{365})^{3650}=1832.09$
$P=1500,~~n=continuous,~~t=10,~~r=0.02=2$%:
$A=1500e^{0.02\times10}=1832.10$
