A customer deposits $2000 in a savings account that pays 5.2% interest compounded quarterly. How much money will the customer have in the account after 2 yr? After 5 yr?

Respuesta :

To solve this we are going to use the compound interest formula: 
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
where
[tex]A[/tex] is the amount after [tex]t[/tex] years 
[tex]P[/tex] is the initial amount 
[tex]r[/tex] is the interest rate in decimal form 
[tex]n[/tex] is the number of times the interest is compounded per year
[tex]t[/tex] is the time in years.

First, we are going to convert the interest rate to decimal form by divide the rate by 100%:
[tex] \frac{5.2}{100} =0.052[/tex]
Next, we are going to find [tex]n[/tex]. Since the interest is compounded quarterly, it is compounded 4 times per year, so.
[tex]n=4[/tex]

1. For our problem we know that [tex]t=2[/tex] and [tex]P=2000[/tex]. We also know for our previous calculations that [tex]r=0.052[/tex] and [tex]n=4[/tex]. So lets replace those values in our compound interest formula to find [tex]A[/tex]:
[tex]A=P(1+ \frac{r}{n} )^{nt} [/tex]
[tex]A=2000(1+ \frac{0.052}{4} )^{(2)(4)[/tex]
[tex]A=2000(1+ \frac{0.052}{4})^{8} [/tex]
[tex]A=2217.71[/tex]
We can conclude that after 2 years the customer will have $2,217.71 in his account.

2. We know for our problem that this time [tex]t=5[/tex], the initial investment remains the same, so [tex]p=2000[/tex], and we also know for our previous calculations that [tex]r=0.052[/tex] and [tex]n=4[/tex]. So lets replace those values in our formula one more time:
[tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
[tex]A=2000(1+ \frac{0.052}{4} )^{(4)(5)[/tex]
[tex]A=2000(1+ \frac{0.052}{4} )^{20}[/tex]
[tex]A=2589.52[/tex]
We can conclude that after 5 years the customer will have $2,589.52 in his account.
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