if 3000 dollars is invested in a bank account at an interest rate of 6 percent per year find the amount in the bank after 10 years if interest is compounded quarterly
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
P=3,000
r=6%=6/100=0.06
t=10 years
n=4
substtute in the formula above
[tex]\begin{gathered} A=3,000(1+\frac{0.06}{4})^{4\cdot10} \\ \\ A=3,000(\frac{4.06}{4})^{40} \\ \\ A=\$5,442.06 \end{gathered}[/tex]Part 2
compounded monthly
we have
P=3,000
r=6%=6/100=0.06
t=10 years
n=12
substtute in the formula above
[tex]\begin{gathered} A=3,000(1+\frac{0.06}{12})^{12\cdot10} \\ \\ A=3,000(\frac{12.06}{12})^{120} \\ \\ A=\$5,458.19 \end{gathered}[/tex]Part 3
continuously
we know that
The formula to calculate continuously compounded interest is equal to
[tex]A=P(e)^{rt}[/tex]
[tex]A=P\mleft(e\mright)^{\mleft\{rt\mright\}}[/tex]where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
e is the mathematical constant number
we have
P=3,000
r=6%=6/100=0.06
t=10 years
substitute
[tex]\begin{gathered} A=3,000(e)^{\{0.06\cdot10\}} \\ A=\$5,466.36 \end{gathered}[/tex]