// Numbas version: finer_feedback_settings
{"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": "<p>Solving for a geometric series</p>"}, "ungrouped_variables": ["n", "s1", "s2", "r", "a"], "parts": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "prompt": "<p>Determine the value of the common ratio. &nbsp; &nbsp;\\(r\\) =&nbsp;[[0]]</p>\n<p>Calculate the value of the first term. &nbsp; &nbsp;\\(a\\) =&nbsp;[[1]]</p>", "variableReplacements": [], "gaps": [{"notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReduced": false, "strictPrecision": false, "showCorrectAnswer": true, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacements": [], "minValue": "{r}", "scripts": {}, "marks": 1, "showFeedbackIcon": true, "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": 0, "precisionType": "dp", "type": "numberentry", "correctAnswerFraction": false, "maxValue": "{r}", "precision": "2"}, {"notationStyles": ["plain", "en", "si-en"], "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision.", "mustBeReduced": false, "strictPrecision": false, "showCorrectAnswer": true, "allowFractions": false, "correctAnswerStyle": "plain", "variableReplacements": [], "minValue": "{a}", "scripts": {}, "marks": 1, "showFeedbackIcon": true, "showPrecisionHint": true, "mustBeReducedPC": 0, "precisionPartialCredit": 0, "precisionType": "dp", "type": "numberentry", "correctAnswerFraction": false, "maxValue": "{a}", "precision": "2"}], "marks": 0}], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"n": {"name": "n", "description": "", "definition": "random(5..12#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "s1": {"name": "s1", "description": "", "definition": "random(12..50#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "r": {"name": "r", "description": "", "definition": "(1-s1/s2)^(1/{n})", "templateType": "anything", "group": "Ungrouped variables"}, "a": {"name": "a", "description": "", "definition": "{s2}*(1-{r})", "templateType": "anything", "group": "Ungrouped variables"}, "s2": {"name": "s2", "description": "", "definition": "2*{s1}-{n}", "templateType": "anything", "group": "Ungrouped variables"}}, "extensions": [], "type": "question", "preamble": {"js": "", "css": ""}, "advice": "<p>\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s1}\\)</p>\n<p>\\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)</p>\n<p>If we divide one by the other we get:</p>\n<p>\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{\\frac{a(1-r^{\\simplify{n}})}{1-r}}{\\frac{a}{1-r}}=\\frac{\\var{s1}}{\\var{s2}}\\)</p>\n<p>\\(\\frac{S_{\\simplify{n}}}{S_{\\infty}}=\\frac{a(1-r^{\\simplify{n}})}{1-r}*\\frac{1-r}{a}=\\frac{\\var{s1}}{\\var{s2}}\\)</p>\n<p>\\(1-r^{{n}}=\\frac{\\var{s1}}{\\var{s2}}\\)</p>\n<p>\\(1-\\frac{\\var{s1}}{\\var{s2}}=r^{{n}}\\)</p>\n<p>In this example \\(n=\\var{n}\\)</p>\n<p>\\(r^{\\var{n}}=\\simplify{({s2}-{s1})/{s2}}\\)</p>\n<p>\\(r=(\\simplify{({s2}-{s1})/{s2}})^{1/\\var{n}}\\)</p>\n<p>\\(r=\\simplify{(({s2}-{s1})/{s1})^{1/{n}}}=\\var{r}\\)</p>\n<p>Recall \\(S_{\\infty}=\\frac{a}{1-r}=\\var{s2}\\)</p>\n<p>\\(a=\\var{s2}*(1-{r})\\)</p>\n<p>Inserting the value for \\(r\\) in this equation gives</p>\n<p>\\(a=\\var{s2}*\\simplify{(1-{r})}\\)</p>\n<p>\\(a=\\var{a}\\)</p>\n<p></p>", "statement": "<p>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}\\).</p>", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}