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{"name": "Arithmetic progressions: simultaneous equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Arithmetic progressions: simultaneous equations", "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"css": "", "js": ""}, "functions": {}, "parts": [{"type": "gapfill", "scripts": {}, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0, "showFeedbackIcon": true, "showCorrectAnswer": true, "prompt": "<p>Calculate the value of the common difference. &nbsp; \\(d\\) =&nbsp;[[0]]</p>\n<p>Calculate the value of the first term of the series. &nbsp;\\(a\\) =&nbsp;[[1]]</p>", "gaps": [{"precisionType": "dp", "scripts": {}, "precision": "1", "minValue": "{d}", "allowFractions": false, "showPrecisionHint": true, "mustBeReducedPC": 0, "mustBeReduced": false, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "marks": 1, "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "type": "numberentry", "maxValue": "{d}", "correctAnswerStyle": "plain", "strictPrecision": false, "showFeedbackIcon": true, "correctAnswerFraction": false}, {"precisionType": "dp", "scripts": {}, "precision": "1", "minValue": "{a}", "allowFractions": false, "showPrecisionHint": true, "mustBeReducedPC": 0, "mustBeReduced": false, "showCorrectAnswer": true, "notationStyles": ["plain", "en", "si-en"], "precisionMessage": "You have not given your answer to the correct precision.", "marks": 1, "precisionPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "type": "numberentry", "maxValue": "{a}", "correctAnswerStyle": "plain", "strictPrecision": false, "showFeedbackIcon": true, "correctAnswerFraction": false}]}], "rulesets": {}, "variables": {"a": {"name": "a", "description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "{T}-({n2}-1)*{d}"}, "n2": {"name": "n2", "description": "", "group": "Ungrouped variables", "templateType": "randrange", "definition": "random(5..14#1)"}, "d": {"name": "d", "description": "", "group": "Ungrouped variables", "templateType": "anything", "definition": "({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))"}, "T": {"name": "T", "description": "", "group": "Ungrouped variables", "templateType": "randrange", "definition": "random(8..16#1)"}, "n1": {"name": "n1", "description": "", "group": "Ungrouped variables", "templateType": "randrange", "definition": "random(3..12#1)"}, "s1": {"name": "s1", "description": "", "group": "Ungrouped variables", "templateType": "randrange", "definition": "random(12..60#1)"}}, "extensions": [], "tags": [], "advice": "<p>Recall the formula for the sum of the first n terms of an arithmetic progression is \\(S_n=\\frac{n}{2}(2a+(n-1)d)\\).</p>\n<p>The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\)</p>\n<p style=\"padding-left: 30px;\">\\(\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\\) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; equation (i)</p>\n<p>The formula for nth term of an arithmetic progression is \\(T_n=a+(n-1)d\\).</p>\n<p>The \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\)</p>\n<p style=\"padding-left: 30px;\">\\(a+\\simplify{{n2}-1}d=\\var{T}\\) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; equation (ii)</p>\n<p>Here we have two simultaneous equations. We can eliminate the \\(a\\) term.</p>\n<p style=\"padding-left: 30px;\">\\(\\var{n1}a+\\simplify{({n1}-1)*{n1}/2}d=\\var{s1}\\) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;equation (i)</p>\n<p style=\"padding-left: 30px;\">\\(\\var{n1}a+\\simplify{{n1}*({n2}-1)}d=\\simplify{{n1}*{T}}\\) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;equation (ii)*\\(\\var{n1}\\)</p>\n<p style=\"padding-left: 30px;\">\\(\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}d=\\simplify{{s1}-{n1}*{T}}\\)</p>\n<p style=\"padding-left: 30px;\">\\(d=\\frac{\\simplify{{s1}-{n1}*{T}}}{\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}}\\)</p>\n<p style=\"padding-left: 30px;\">\\(d=\\simplify{({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)</p>\n<p>Using this result and equation (ii) we can find the value for \\(a\\)</p>\n<p style=\"padding-left: 30px;\">\\(a+\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}=\\var{T}\\)</p>\n<p style=\"padding-left: 30px;\">\\(a=\\var{T}-\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)</p>\n<p style=\"padding-left: 30px;\">\\(a=\\simplify{{T}-({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)</p>\n<p style=\"padding-left: 30px;\"></p>", "ungrouped_variables": ["n1", "s1", "n2", "T", "d", "a"], "metadata": {"licence": "Creative Commons Attribution-NonCommercial 4.0 International", "description": "<p>Solving arithmetic progressions using simultaneous equations</p>"}, "statement": "<p>The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\) and the \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\).</p>", "variable_groups": [], "type": "question", "contributors": [{"name": "karl walton", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2516/"}]}]}], "contributors": [{"name": "karl walton", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2516/"}]}