Question

David’s bank offers a 36-month Certificate of Deposit (CD) with an APR of 2.25%. Use the compound interest formula to answer the following: a) If P = 2000, what is A(8)? b) Solve the equation A(t) = 4000 for t. c) What principal P should be invested so that the account balance is $2000 in three years?

Answer 1

Using compound interest, it is found that:

a) A(8) = 2389.66

b) t = 31.15

c) P = 1870.85

Compound interest:

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

• A(t) is the amount of money after t years.  
• P is the principal(the initial sum of money).  
• r is the interest rate(as a decimal value).  
• n is the number of times that interest is compounded per year.  
• t is the time in years for which the money is invested or borrowed.

In this problem:

• The APR is of 2.25%, hence $r = 0.0225$.

• No information about the number of compounding per year, hence $n = 1$.

Item a:

$P = 2000$, hence:

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

$A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}$

$A(8) = 2389.66$

Item b:

$A(t) = 4000$, hence:

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

$4000 = 2000\left(1 + \frac{0.0225}{1}\right)^{t}$

$(1.0225)^t = 2$

$\log{(1.0225)^t} = \log{2}$

$t\log{1.0225} = \log{2}$

$t = \frac{\log{2}}{\log{1.0225}}$

$t = 31.15$

Item c:

$A(3) = 2000$, hence:

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

$2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}$

$P = \frac{2000}{(1.0225)^3}$

$P = 1870.85$

A similar problem is given at brainly.com/question/24850750