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Thread: Arithmetic and geometric progression

  1. #1
    Jul 2017

    Arithmetic and geometric progression

    Anyone can help me for question 10?...i have managed to solved question no.9, but i have no idea with no.10....

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  2. #2
    MHF Contributor

    Apr 2005

    Re: Arithmetic and geometric progression

    That's very hard to read- especially side ways! Problem 10 is
    Tin ore is extracted from a certain mine. During the first year of production, a yield of 8000 kg of tin ore was obtained. With the increasing difficulty of mining, the production of tin ore production in each subsequent year show a decrease of 10% on the previous years production. Assuming that mining continues in the same way for an indefinite time, calculate the maximum amount of tin ore that could probably be extracted.
    This an infinite geometric series with initial amount 8000 and "common ratio" 1- 0.1= .9. The formula for the sum of a geometric series is $\displaystyle \sum_{i= 0}^\infty ar^i= \frac{a}{1- r}$.

    For economic reasons, mining is abandoned once the annual output of tin ore falls below 1000 kg.
    (a) The number of years the mine is in production.
    The "nth" years production is $\displaystyle (8000)(0.9)^n$ Find n so that is less than 1000. You may need to use a logarithm.

    (b) the total production of tin ore in this time.
    The finite sum of a geometric series is $\displaystyle \sum_{i=0}^n at^i= \frac{a(1- r^{n+1}}{1- r}$. Use the same a and r as in the first question and the "n" from (a).

    (c) the percentage of tin ore not extracted from this mine.
    The first question, with the infinite series, gave the total amount of tin ore. (b) gave the amount that could be extracted. Subtract the second from the first to get the amount of tin ore left in the mine. Then divide by the total amount to get the percentage.
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