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Math Help - series and sequences

  1. #1
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    series and sequences

    prospectors are drilling for oil. The cost of drilling to a depth of 50m is 500. To drill a further 50m costs 640 and, hence, the total cost of drilling to a depth of 100m is 1140. Each subsequent extra depth of 50m costs 140 more to drill than the previous 50m.
    the total sum of money available for drilling is 76000. Find, to the nearest 50m, the freatest depth that can be drilled.
    a=500 d=140 i'm not sure what n=....
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  2. #2
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    Hello, girlpower1991!

    Prospectors are drilling for oil.
    The cost of drilling to a depth of 50m is $500.
    To drill a further 50m costs $640
    and, hence, the total cost of drilling to a depth of 100m is $1140.

    Each subsequent extra depth of 50m costs $140 more to drill than the previous 50m.
    The total sum of money available for drilling is $76,000.
    Find, to the nearest 50m, the greatest depth that can be drilled.
    I had to baby-talk mny way through it . . .

    \begin{array}{cccc}\text{Level} & \text{Cost per level} & \text{Total cost} \\ \hline 1 & 500 \\ & & 500 \\ 2 & 640 \\ & & 1140 \\ 3 & 780 \\ & & 1920 \\ 4 & 920 \\ & & 2840 \end{array}
    . . \begin{array}{ccccc} 5 & \qquad\;\; 1060 \\ & & \qquad\quad\; 3900 \end{array}

    We have a Cost sequence: . 500, 1140, 1920, 2840, 3900, \hdots


    Take the difference of consecutive terms,
    . . then take differences of the differences, etc.

    \begin{array}{cccccccccc} \text{Sequence} & 500 && 1140 && 1920 && 2840 && 3900 \\ \text{1st diff.} & & 640 & & 780 & & 920 & & 1060 \\ \text{2nd diff.} & & & 140 & & 140 & & 140  \end{array}


    The second differences are constant.
    . . Hence, the function is of the second degree . . . a quadratic.

    The general quadratic function is: . f(n) \;=\;an^2 + bn + c


    Plug in the first three values of the sequence:

    . . \begin{array}{ccccc} f(1) = 500\!: & a + b + c &=& 500 & [1] \\<br />
f(2) = 1140\!: & 4a + 2b + c &=& 1140 & [2] \\<br />
f(3) = 1920\!: & 9a + 3b + c &=& 1920 & [3] \end{array}


    \begin{array}{ccccc}\text{Subtract [2] - [1]:} & 3a + b &=& 640 & [4] \\<br />
\text{Subtract [3] - [2]:} & 5a + b &=& 780 & [5] \end{array}

    \begin{array}{cccccccc}\text{Subtract [5] - [4]:} & 2a \:=\: 140 & \Rightarrow &\boxed{a \:=\:70}\end{array}

    \begin{array}{ccccccc}\text{Substitute into [4]:} & 3(70) + b \:=\:640 & \Rightarrow &\boxed{b \:=\:430} \end{array}

    \begin{array}{cccccc}\text{Substitute into [1]:} & 70 + 430 + c &=& 500 & \Rightarrow &\boxed{c \:=\:0} \end{array}

    Hence: . \boxed{f(n) \;=\;70n^2 + 430n}


    We have: . 70n^2 + 430n \;=\;76,\!000 \quad\Rightarrow\quad 7n^2 + 43n - 7600 \;=\;0

    Quadratic Formula: .  n \;=\;\frac{-43\pm\sqrt{43^2 + 4(7)(7600)}}{14} \;=\; <br />
\frac{-43 \pm\sqrt{214,649}}{14}

    The positive root is: . n \;=\;30.02159122 \;\approx\;30 levels


    The greatest depth is: .  30 \times 50 \;=\;1500 m.

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