Total amount after 30 years is given by :
[tex]A=\dfrac{d((1+\dfrac{r}{k})^{Nk}-1}{\dfrac{r}{k}}[/tex]
[tex]A=\dfrac{200((1+\dfrac{0.03}{12})^{30\times 12}-1}{\dfrac{0.03}{12}}\\\\\\A=\$116547.38[/tex]
Now, total money deposited, T = 200×30×12 = $72000 .
Therefore, we have around $116547.38 in our account and total interest earned is $72000 .
Hence, this is the required solution.
