1. ## Geometric progression.

A precious metal is extracted from a min. In the first year of operation, 2000kg of the metal was extracted. In each succeeding year, the amount extracted was 90% of the previous year's amount.

i)Find the amount of metal extracted in the 10th year of operation.
ii)the total amount of metal extracted in the first 20 years of operation.
iii)the total amount of metal that would be extracted over a very long period of time.

I have no idea how to answer these 3 questions. To be honest, I very suck at this topic. Never learned it before. I was not in school when my tutor taught this :S Anyone knows a good site for this topic.

2. Originally Posted by Berkshire91
A precious metal is extracted from a min. In the first year of operation, 2000kg of the metal was extracted. In each succeeding year, the amount extracted was 90% of the previous year's amount.

i)Find the amount of metal extracted in the 10th year of operation.
ii)the total amount of metal extracted in the first 20 years of operation.
iii)the total amount of metal that would be extracted over a very long period of time.

I have no idea how to answer these 3 questions. To be honest, I very suck at this topic. Never learned it before. I was not in school when my tutor taught this :S Anyone knows a good site for this topic.
get caught up ...

http://www.usna.edu/MathDept/website...21_geomser.pdf

3. ${a_1},{a_2},...{a_n}...$ - geometric progression
${a_1} = 2000$
r-common ratio
r=0.9 (It is yours 90 %)

$\begin{array}{l}
{a_2} = {a_1} \times r \\
{a_3} = {a_1} \times {r^2} \\
\end{array}
$

...
${a_{10}} = {a_1} \times {r^9}
$

Substitution ${a_1}$ and r
1) ${a_{10}} = 2000 \times {0.9^9}$
2) For calculation total amount in the first 20 years use this formula:
$\sum\limits_{k = 0}^n {{a_1}{r^k}} = \frac{{{a_1}({r^{n + 1}} - 1)}}{{r - 1}}
$

3) $\sum\limits_{k = 0}^\infty {{a_1}{r^k}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {{a_1}{r^k}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}({r^{n + 1}} - 1)}}{{r - 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{r^{n + 1}}}}{{r - 1}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}}}{{r - 1}} = \frac{{{a_1}}}{{1 - r}}{\kern 1pt} {\kern 1pt}
$
(Sience ${r^{n + 1}} \to 0{\kern 1pt}
$
) then $\mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{r^{n + 1}}}}{{r - 1}} = 0
$

Substitution ${a_1}$ and r
$\sum\limits_{k = 0}^\infty {{a_1}{r^k}} = \frac{{2000}}{{1 - 0.9}} = 20000{\kern 1pt} {\kern 1pt} (kg)
$