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Linda runs a smart phone app business, and in her first year she had net earnings of \($20\thinspace 000\). Since then, her net earnings have increased by \(5\%\) each year.

  1. Write out how much money Linda earned each year, for her first five years in business.

    In the first year, Linda earned \($20\thinspace 000\).
    In the second year, Linda earned \($20\thinspace 000 + 0.05($20\thinspace 000) = $21\thinspace 000\).
    In the third year, Linda earned \($21\thinspace 000 + 0.05($21\thinspace 000) = $22\thinspace 050\).
    In the fourth year, Linda earned \($22\thinspace 050 + 0.05($22\thinspace 050) = $23\thinspace 152.50\).
    In the fifth year, Linda earned \($23\thinspace 152.50 + 0.05($23\thinspace 152.50) = $24\thinspace 310.125 \buildrel\textstyle.\over =  $24\thinspace 310.13\).

  2. Justify that this is a geometric sequence and write the general term.

    The ratio for each pair of consecutive terms is
    \(\begin{array}{l}
     r = \frac{21\thinspace 000}{20\thinspace 000} = 1.05 \\
     r = \frac{22\thinspace 050}{21\thinspace 000} = 1.05 \\
     r = \frac{23\thinspace 152.50}{22\thinspace 050} = 1.05 \\
     r = \frac{24\thinspace 310.125}{23\thinspace 152.50} = 1.05 \\
     \end{array}\)


    Notice that the values are getting larger each year. This indicates that \(r \) is greater than 1.
    Because the ratio is constant, this is a geometric sequence.

    The general term is \(t_n = 20\thinspace 000\left( {1.05} \right)^{n - 1} \).

  3. How much money has Linda earned through the business after 5 years? 10 years?

    This is asking for the sum of geometric series.
    \(\begin{align}
     S_n &= \frac{t_1 \left( {r^n  - 1} \right)}{r - 1} \\
     S_5 &= \frac{20\thinspace 000\left[ {\left( {1.05} \right)^5 - 1} \right]}{1.05 - 1} \\
     S_5 &= 110\thinspace 512.625 \\
     S_5 &= \$110\thinspace 512.63 \\
     \end{align}\)
    \(\begin{align}
     S_n &= \frac{t_1 \left( {r^n  - 1} \right)}{r - 1} \\
     S_{10} &= \frac{20\thinspace 000\left[ {\left( {1.05} \right)^{10} - 1} \right]}{1.05 - 1} \\
     S_{10} &= 251\thinspace 557.850\thinspace 7 \\
     S_{10} &= \$251\thinspace 557.85 \\
     \end{align}\)

  4. Linda’s goal is to net one million dollars through the business. How many years will it take for her to accomplish this goal?

    \(
    \begin{align}
     S_n &= \frac{t_1 \left( {r^n - 1} \right)}{r - 1} \\
     1\thinspace 000\thinspace 000 &= \frac{20\thinspace 000\left[ {\left( {1.05} \right)^n - 1} \right]}{1.05 - 1} \\
     2.5 &= \left( {1.05} \right)^n  - 1 \\
     3.5 &= \left( {1.05} \right)^n  \\
     \end{align}
    \)
    Use the process of guess and check to find the value of \(n \) that will work.

    \(1.05^{24}  = 3.225...\)         too small
    \(1.05^{30}  = 4.321...\)         too large
    \(1.05^{28}  = 3.920...\)         too large
    \(1.05^{25}  = 3.386...\)         getting closer
    \(1.05^{26}  = 3.555...\)         large enough!

    It will take Linda between 25 and 26 years to net one million dollars through her business.

  5. What assumption(s) did you make in answering c. and d.?

    Linda's business continues to do well enough for her net earnings to increase by \(5\%\).