// Numbas version: exam_results_page_options {"name": "Andrew's copy of Solving for a geometric series", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"preamble": {"css": "", "js": ""}, "tags": [], "functions": {}, "parts": [{"marks": 0, "gaps": [{"marks": 1, "variableReplacements": [], "correctAnswerFraction": false, "minValue": "{r}", "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "maxValue": "{r}", "allowFractions": false, "precisionType": "dp", "type": "numberentry", "precision": "2", "scripts": {}, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "showPrecisionHint": true, "strictPrecision": false}, {"marks": 1, "variableReplacements": [], "correctAnswerFraction": false, "minValue": "{a}", "precisionMessage": "You have not given your answer to the correct precision.", "precisionPartialCredit": 0, "maxValue": "{a}", "allowFractions": false, "precisionType": "dp", "type": "numberentry", "precision": "2", "scripts": {}, "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst", "showPrecisionHint": true, "strictPrecision": false}], "scripts": {}, "showCorrectAnswer": true, "prompt": "

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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The sum of the first \\(\\var{n}\\) terms of a geometric series is \\(\\var{s_1}\\) and the sum of the first \\(\\simplify{2*{n}}\\) terms is \\(\\var{s_2}\\).

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Solving for a geometric series

"}, "variables": {"r": {"definition": "({s_2}/{s_1}-1)^(1/{n})", "description": "", "templateType": "anything", "name": "r", "group": "Ungrouped variables"}, "n": {"definition": "random(2..6#1)", "description": "", "templateType": "randrange", "name": "n", "group": "Ungrouped variables"}, "a": {"definition": "{s_1}*(1-{r})/(1-{r}^{n})", "description": "", "templateType": "anything", "name": "a", "group": "Ungrouped variables"}, "s_2": {"definition": "random(36..50#1)", "description": "", "templateType": "randrange", "name": "s_2", "group": "Ungrouped variables"}, "s_1": {"definition": "random(12..36#1)", "description": "", "templateType": "randrange", "name": "s_1", "group": "Ungrouped variables"}}, "name": "Andrew's copy of Solving for a geometric series", "advice": "

\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

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\\(S_{\\simplify{2*{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}=\\var{s_2}\\)

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If we divide one by the other we get:

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\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}}{\\frac{a(1-r^{\\var{n}})}{1-r}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}*\\frac{1-r}{a(1-r^{\\var{n}})}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{1-r^{\\simplify{2*{n}}}}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{(1-r^\\var{n})(1+r^{\\var{n}})}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(1+r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}-1\\)

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\\(r^{\\var{n}}=\\simplify{{s_2}/{s_1}-1}\\)

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\\(r=\\simplify{(({s_2})/{s_1}-1)^{1/{n}}}\\)

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\\(r=\\simplify{(({s_2}-{s_1})/{s_1})^{1/{n}}}=\\var{r}\\)

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Recall \\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

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\\(a=\\frac{\\var{s_1}*(1-{r})}{1-r^{\\var{n}}}\\)

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

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\\(a=\\frac{\\var{s_1}*(\\simplify{(1-{r})})}{\\simplify{{1-r^{{n}}}}}\\)

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

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", "type": "question", "contributors": [{"name": "Andrew Dunbar", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/770/"}]}]}], "contributors": [{"name": "Andrew Dunbar", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/770/"}]}