// Numbas version: exam_results_page_options {"name": "Solving for a geometric series #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variable_groups": [], "tags": [], "name": "Solving for a geometric series #2", "functions": {}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "

Solving for a geometric series

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Determine the value of the common ratio.    \\(r\\) = [[0]]

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Calculate the value of the first term.    \\(a\\) = [[1]]

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\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s1}\\)

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\\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)

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If we divide one by the other we get:

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\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{\\frac{a(1-r^{\\simplify{n}})}{1-r}}{\\frac{a}{1-r}}=\\frac{\\var{s1}}{\\var{s2}}\\)

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\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{a(1-r^{\\simplify{n}})}{1-r}*\\frac{1-r}{a}=\\frac{\\var{s1}}{\\var{s2}}\\)

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\\(1-r^{{n}}=\\frac{\\var{s1}}{\\var{s2}}\\)

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\\(1-\\frac{\\var{s1}}{\\var{s2}}=r^{{n}}\\)

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In this example \\(n=\\var{n}\\)

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\\(r^{\\var{n}}=\\simplify{({s2}-{s1})/{s2}}\\)

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\\(r=(\\simplify{({s2}-{s1})/{s2}})^{1/\\var{n}}\\)

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\\(r=\\simplify{(({s2}-{s1})/{s1})^{1/{n}}}=\\var{r}\\)

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Recall \\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)

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\\(a=\\var{s2}*(1-{r})\\)

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Inserting the value for \\(r\\) in this equation gives

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\\(a=\\var{s2}*\\simplify{(1-{r})}\\)

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\\(a=\\var{a}\\)

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The sum of the first \\(\\var{n}\\) terms of a geometric series is \\(\\var{s1}\\) and the sum to infinity of the series is \\(\\var{s2}\\).

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