// Numbas version: exam_results_page_options {"name": "Simon's copy of Solving for a geometric series #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"metadata": {"description": "

Solving for a geometric series

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "

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}\$$.

", "ungrouped_variables": ["n", "s1", "s2", "r", "a"], "variable_groups": [], "variables": {"s2": {"description": "", "definition": "2*{s1}-{n}", "group": "Ungrouped variables", "name": "s2", "templateType": "anything"}, "s1": {"description": "", "definition": "random(12..50#1)", "group": "Ungrouped variables", "name": "s1", "templateType": "randrange"}, "a": {"description": "", "definition": "{s2}*(1-{r})", "group": "Ungrouped variables", "name": "a", "templateType": "anything"}, "r": {"description": "", "definition": "(1-s1/s2)^(1/{n})", "group": "Ungrouped variables", "name": "r", "templateType": "anything"}, "n": {"description": "", "definition": "random(5..13#2)", "group": "Ungrouped variables", "name": "n", "templateType": "randrange"}}, "parts": [{"marks": 0, "gaps": [{"precisionMessage": "You have not given your answer to the correct precision.", "variableReplacements": [], "scripts": {}, "correctAnswerStyle": "plain", "unitTests": [], "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "precisionType": "dp", "precision": "3", "marks": 1, "precisionPartialCredit": 0, "correctAnswerFraction": false, "mustBeReducedPC": 0, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "{r}", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "strictPrecision": false, "showPrecisionHint": true, "maxValue": "{r}", "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": ""}, {"precisionMessage": "You have not given your answer to the correct precision.", "variableReplacements": [], "scripts": {}, "correctAnswerStyle": "plain", "unitTests": [], "notationStyles": ["plain", "en", "si-en"], "showFeedbackIcon": true, "precisionType": "dp", "precision": "3", "marks": 1, "precisionPartialCredit": 0, "correctAnswerFraction": false, "mustBeReducedPC": 0, "mustBeReduced": false, "type": "numberentry", "showCorrectAnswer": true, "minValue": "{a}", "variableReplacementStrategy": "originalfirst", "allowFractions": false, "strictPrecision": false, "showPrecisionHint": true, "maxValue": "{a}", "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": ""}], "prompt": "

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

\n

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

", "variableReplacements": [], "type": "gapfill", "scripts": {}, "unitTests": [], "showCorrectAnswer": true, "sortAnswers": false, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": ""}], "tags": [], "rulesets": {}, "name": "Simon's copy of Solving for a geometric series #2", "functions": {}, "advice": "

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

\n

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

\n

If we divide one by the other we get:

\n

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

\n

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

\n

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

\n

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

\n

In this example \$$n=\\var{n}\$$

\n

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

\n

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

\n

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

\n

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

\n

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

\n

Inserting the value for \$$r\$$ in this equation gives

\n

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

\n

\$$a=\\var{a}\$$

\n

", "preamble": {"css": "", "js": ""}, "extensions": [], "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}