Question

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

Answer 1

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

$A=P(1+\frac{r}{n})^{nt}$

$A=P(1+\frac{r}{n})^{nt}$

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

$\begin{gathered} A=3,000(1+\frac{0.06}{4})^{4\cdot10} \\ \\ A=3,000(\frac{4.06}{4})^{40} \\ \\ A=\ 5,442.06 \end{gathered}$

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

$\begin{gathered} A=3,000(1+\frac{0.06}{12})^{12\cdot10} \\ \\ A=3,000(\frac{12.06}{12})^{120} \\ \\ A=\ 5,458.19 \end{gathered}$

Part 3

continuously

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

$A=P\mleft(e\mright)^{\mleft\{rt\mright\}}$

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

$\begin{gathered} A=3,000(e)^{\{0.06\cdot10\}} \\ A=\ 5,466.36 \end{gathered}$