Answer
(a) $2\frac{1}{2}\%$ per year, compounded semiannually is the best investment.
Work Step by Step
*A quick review* $A = P(1+\frac{r}{n})^{nt}$
$n$ - number of times investment is compounded per year
$t$ - number of years
$r$ - interest rate per year (In decimal form)
$P$ - principal
$A$ - amount of money after $t$ years
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To determine which of them would be the best investment we have to calculate each of them using the compound interest formula.
For principal we can take any arbitrary number, but for easier calculation, let's say $P=\$1000$ and consider that $t=1$
(a) $r=2\frac{1}{2}\%=0.025$ Compounded semiannually, $n=2$ $A=1000(1+\frac{0.025}{2})^{2\times1}=1000\times 1.0125^2=\$1025.16$
(b) $r=2\frac{1}{4}\%=0.0225$ Compounded semiannually, $n=12$ $A=1000(1+\frac{0.0225}{12})^{12\times1}=1000\times 1.001875^{12}\approx\$1022.73$
(c) $r=2%=0.02$ Compounded continuously.
This time we have to use different formula which is used to calculate continuously compounded interest:
$A= Pe^{rt}$ $A=1000e^{0.02\times 1}=\$1020.2$
As we can clearly see from the results, the best investment is in (a) case.
