Multimedia


Example 1


Luke has a landscaping business, and offers lawn mowing services. At the start of his second season, one of his lawn mowers has \(1\thinspace 000\) hours on it. At the start of his fifth season, this same mower has \(4\thinspace 200\) hours on it.
  1. How many hours does Luke put on his machine each season, assuming it follows an arithmetic series?

    Step 1: Write what is given.

    In this problem, \(d \) represents the number of hours Luke puts on the machine each season, so we need to find \(d \). The first term is also unknown. Start by writing down what is given, producing two equations with two unknowns each.

    Equation I
    \(\begin{align}
    S_2 &= 1\thinspace 000 \\
    n &= 2 \\
    d &= ? \\
    t_1 &= ? \\
    \end{align}
    \)
    \[\begin{align}
     S_n &= \frac{n}{2}\left[ {2t_1 + (n - 1)d} \right] \\
     1\thinspace 000 &= \frac{2}{2}\left[ {2t_1 + \left( {2 - 1} \right)d} \right] \\
     1\thinspace 000 &= 2t_1 + d \\
     \end{align}
    \]

    Equation II
    \(\begin{align}
    S_2 &= 4\thinspace 200\\
    n &= 5 \\
    d &= ? \\
    t_1 &= ? \\
    \end{align}
    \)
    \[
    \begin{align}
     S_n &= \frac{n}{2}\left[ {2t_1 + (n - 1)d} \right] \\
     4\thinspace 200 &= \frac{5}{2}\left[ {2t_1 + \left( {5 - 1} \right)d} \right] \\
     4\thinspace 200 &= 2.5\left[ {2t_1 + 4d} \right] \\
     1\thinspace 680& = 2t_1 + 4d \\
     \end{align}
    \]

    Step 2: Set up a system of linear equations, and solve for \(d\).

    Subtract Equation I from Equation II.

    \[\begin{align}
    1\thinspace 680 &= \cancel{t_1} + &4d \\
    -(1\thinspace 000 &= \cancel{t_1} + &d)  \\
    \hline {} \\
    680 &= &3d \\ 
    226.666... &= &d \\
    227 &\buildrel\textstyle.\over= &d \\ 
    \end{align}\]
    Luke increases the amount of hours of machine use added by approximately \(227\) hours per year.

  2. How many hours did the mower have on it when Luke first bought it (at the start of his first season)?

    Use one of the equations to find \(t_1 \).
    \(\begin{align}
     1\thinspace 680 &= 2t_1 + 4d \\
     1\thinspace 680 &= 2t_1 + 4\left( {226.666...} \right) \\
     773.333... &= 2t_1  \\
     386.666... &= t_1  \\
     \end{align}
    \)
    When Luke first started the business, the mower had approximately \(386\) hours on it. Note: In this case, you must round down because the mower’s gauge would not have changed to \(387\) hours yet.

  3. Predict how many hours Luke's machine will have on it at the beginning of the sixth season.

    Use the longer version of the arithmetic series formula to determine \(S_6 \) .
    \[\begin{align}
     S_n &= \frac{n}{2}\left[ {2t_1  + (n - 1)d} \right] \\
     S_6 &= \frac{6}{2}\left[ {2\left( {386.666...} \right) + \left( {6 - 1} \right)226.666...} \right] \\
     S_6 &= 3\left( {1\thinspace 906.666...} \right) \\
     S_6 &= 5\thinspace 719.999... \\
     S_6 &\buildrel\textstyle.\over= 5\thinspace 720 \\
     \end{align}\]
    At the start of the sixth season, the mower will have approximately \(5\thinspace 720\) hours on it.

    Note that you do not round until the very end of the question.


  4. What assumption did you make in order to answer part c.?

    The assumption is that Luke's business is growing at an arithmetic rate, which means the number of hours increases by the same amount each season.