# Word Problem Sequences and Series

#### helpmeddddd

A Mining Company was founded in 1894 and the mineâ€™s initial production was the extraction of 100kg/year of silver. Each following year saw a steady increase of 60kg/year until the silver production peaked at 700 kg/year. Production remained at this level until 1914, when an event caused the mine to close indefinitely. The company then stopped its mining operations on the claim during the First World War (1914â€“1918). In 1920, the company opened up a new second mine which, in the first year, saw a silver yield of 700kg/year. However, with every subsequent year, production decreased by 8%. For economic reasons, the new mine had to close when the production fell below 300kg/year.

1. When did the first mine reach the peak production of 700 kg/year? I figured it out to be 1904, but I didn't use the formula. How should this be worked out?

3. Compare the production of both mines over the first five years. Find the total production figures of the two mines until that ceased operation. which of the two mines was more successful? I think I got 5 years.

4. Calculate the percentage of gold remaining in the ground after all mining operations ceased. No idea.

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#### DarnItJimImAnEngineer

1. Counting years is a valid approach in this case, but a formula would be helpful for longer periods of time, so...
Let $A_n$ be the production rate in year (AD 1893 + n).
$A_1$ = 100 kg/yr
$A_n = A_{n-1} +$60 kg/yr for $n \in {2, 3, 4, â€¦, N}$, where $A_N =$ 700 kg/yr.
If you plot A vs n, you should be able to convince yourself that
$A_n = (100 ~kg/yr) + (60 ~kg/yr)\cdot (n-1)$ within this range.
$A_N = 700 ~kg/yr \rightarrow N = 11$.

2. Let $B_n$ be the production rate of the second mine in year (AD 1919 + n).
$B_1$ = 700 kg/yr
An 8 % decrease each year corresponds to
$B_n = 0.92\cdot B_{n-1}$ for $n \in {2, 3, 4, â€¦, M}$, where $B_M <$ 300 kg/yr.
This is the same as
$B_n = (700 ~kg/yr)\cdot (0.92)^{n-1}$
Finding M:
$300 \leq 700(0.92^{M-1}) \rightarrow 0.92^{M-1} \geq 3/7 \rightarrow M-1 \leq log_{0.92}(3/7) = \frac{ln(3/7)}{ln(0.92)} = 10.2 \rightarrow M = 12$.
In year 11 (1930), it produced 304 kg. In year 12 (1931), it would have produced about 280 kg, and been shut down. With farming production and the national economy down, the fired workers would move west trying to find any kind of a job, or roam around as hobos, while the company founders would make sure their life insurance policies were up to date and take a nice 3:14 nap on the tracks for the 3:15 train.

3. By summing up the $A_n$ and $B_n$ terms over the years of validity, you can find the total amount of silver mined. However, since no information was given about the amount of silver in the mines to begin with, you can't calculate a percentage.

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#### helpmeddddd

The below shows the second mine was more successful.

Is this the right way to do/ another way to do it?

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#### DarnItJimImAnEngineer

Absolutely.

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#### helpmeddddd

Could I have used any of these formulas?

#### DarnItJimImAnEngineer

The 1st and 3rd one look awfully familiar. Look at what I wrote earlier.

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#### skipjack

Forum Staff
A Mining Company was founded in 1894 and the mineâ€™s initial production was the extraction of 100kg/year of silver.
This is unclear. Does it mean that the mine's initial rate of production of silver was 100kg/year or that the mine's total production in its first 12 months was 100kg?

#### DarnItJimImAnEngineer

I think they're simplifying it to production per year (or the yearly average of the production rate). That's probably a more realistic model than treating production as a continuous function, anyway.

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#### helpmeddddd

Why would they give you a question you can't answer?

#### tahirimanov19

It is simple arithmetic series.
$$\displaystyle a_n = a_{n-1} +d = a_1 + (n-1)d$$

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