// Numbas version: exam_results_page_options {"name": "Geometric series - practical application 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["dur", "n", "m1", "m2", "m", "dayn"], "name": "Geometric series - practical application 2", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

The $n$th term in a geometric progression is given by $a_1 \\times r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio. 

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We are given that the first term is $\\var{n}$ and the $\\var{dayn}$th term is $\\simplify{{n}*{m}^({dayn}-1)}$

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Hence we know that $r^{\\var{dayn}-1} = \\frac{\\simplify{{n}*{m}^({dayn}-1)}}{\\var{n}} = \\simplify{{m}^({dayn}-1)}$

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So we find that $r=\\var{m}$

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If the pump were to extract air indefinitely, we look to the formula $S_n = \\frac{a_1}{1-r}$ to find $S_n = \\frac{\\var{n}}{1-\\var{m}} = \\simplify{{n}/(1-{m})}$

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Determine the common ratio of the geometric series

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Calculate the total amount of air that could be extracted from the bottle if the pump were to extract air indefinitely

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A pump is used to extract air from a bottle. The first operation of the pump extracts $\\var{n}$ cm$^3$ of air and subsequent extractions follow a geometric series. Extraction number $\\var{dayn}$ of the pump extracts $\\simplify{{n}*{m}^({dayn}-1)}$ cm$^3$ of air.

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