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?

1 Answer

  • Using compound interest, it is found that:

    a) A(8) = 2389.66

    b) t = 31.15

    c) P = 1870.85

    Compound interest:

    [tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

    • 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 [tex]r = 0.0225[/tex].
    • No information about the number of compounding per year, hence [tex]n = 1[/tex].

    Item a:

    [tex]P = 2000[/tex], hence:

    [tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

    [tex]A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}[/tex]

    [tex]A(8) = 2389.66[/tex]

    Item b:

    [tex]A(t) = 4000[/tex], hence:

    [tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

    [tex]4000 = 2000\left(1 + \frac{0.0225}{1}\right)^{t}[/tex]

    [tex](1.0225)^t = 2[/tex]

    [tex]\log{(1.0225)^t} = \log{2}[/tex]

    [tex]t\log{1.0225} = \log{2}[/tex]

    [tex]t = \frac{\log{2}}{\log{1.0225}}[/tex]

    [tex]t = 31.15[/tex]

    Item c:

    [tex]A(3) = 2000[/tex], hence:

    [tex]A(t) = P\left(1 + \frac{r}{n}\right)^{nt}[/tex]

    [tex]2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}[/tex]

    [tex]P = \frac{2000}{(1.0225)^3}[/tex]

    [tex]P = 1870.85[/tex]

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