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Thread: Geometric progression

  1. #1
    Junior Member
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    Geometric progression

    Each stroke of a pump removes 8.2% of the remaining air from a container. What percent of the air remains after 50 strokes?

    I originally thought this was simple, use the formula $\displaystyle a_n=a_1r^{n-1}$
    where
    $\displaystyle a_1=1$
    $\displaystyle r=0.918$
    and $\displaystyle n=50$

    substituting in these values i get
    $\displaystyle a_{50}=1*0.918 ^{50-1}$
    $\displaystyle a_{50}=0.015111$
    and therefore ~1.51%
    however, the solution in my textbook shows the intermediary step to be
    $\displaystyle a_{50}=1*0.918 ^{50}$
    and final solution as
    $\displaystyle a_{50}=0.01387$

    Which brings me to my main question: Could somebody explain to me why the textbook's solution shows $\displaystyle 0.918^{50}$ instead of $\displaystyle 0.918^{50-1}$? Is it something to do with $\displaystyle a_{1}=1$? a mistake in the book? or something else?

    thanks in advance
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  2. #2
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    Lexington, MA (USA)
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    Hello, mattty!

    You miscounted . . .


    Each stroke of a pump removes 8.2% of the remaining air from a container.
    What percent of the air remains after 50 strokes?

    I originally thought this was simple, use the formula $\displaystyle a_n\:=\:a_1r^{n-1}$ . . . . no
    where: .$\displaystyle a_1=1,\;r=0.918,\;n=50$

    You should be using: .$\displaystyle a_n \:=\:a_1r^{{\color{red}n}}$


    Check out the initial condition.

    When $\displaystyle n = 0,\;a_o \:=\:a_1r^0 \quad\Rightarrow\quad a_o = a_1$

    When no air has been pumped out, the amount is the initial amount, $\displaystyle a_1$

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  3. #3
    Junior Member
    Joined
    Jan 2009
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    that's a forehead smack moment if i ever saw one...
    yeah, i can see now, when $\displaystyle n=1$ 100% of the air has been pumped out
    thanks for pointing out the obvious.... i need that kind of help a lot -_-
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