The sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\) and the sum of the first \\(\\var{n2}\\) terms of an arithmetic progression is \\(\\var{s2}\\)
", "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "tags": [], "variables": {"s1": {"description": "", "name": "s1", "definition": "random(12..60#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "s2": {"description": "", "name": "s2", "definition": "random(48..90#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "a": {"description": "", "name": "a", "definition": "({s1}-{n1}*({n1}-1)*d/2)/{n1}", "templateType": "anything", "group": "Ungrouped variables"}, "d": {"description": "", "name": "d", "definition": "2*({n2}*{s1}-{n1}*{s2})/({n1}*{n2}*({n1}-{n2}))", "templateType": "anything", "group": "Ungrouped variables"}, "n1": {"description": "", "name": "n1", "definition": "random(3..8#1)", "templateType": "randrange", "group": "Ungrouped variables"}, "n2": {"description": "", "name": "n2", "definition": "random(9..15#1)", "templateType": "randrange", "group": "Ungrouped variables"}}, "variable_groups": [], "metadata": {"description": "Solving arithmetic progressions using simultaneous equations
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "parts": [{"marks": 0, "type": "gapfill", "showCorrectAnswer": true, "scripts": {}, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "variableReplacements": [], "prompt": "Calculate the value of the common difference. \\(d\\) = [[0]]
\nCalculate the value of the first term of the series. \\(a\\) = [[1]]
", "gaps": [{"mustBeReducedPC": 0, "marks": 1, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "correctAnswerFraction": false, "variableReplacements": [], "maxValue": "{d}", "minValue": "{d}", "allowFractions": false, "precisionPartialCredit": 0, "type": "numberentry", "correctAnswerStyle": "plain", "showCorrectAnswer": true, "precisionType": "dp", "showPrecisionHint": true, "precision": "1", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision.", "scripts": {}}, {"mustBeReducedPC": 0, "marks": 1, "mustBeReduced": false, "notationStyles": ["plain", "en", "si-en"], "strictPrecision": false, "correctAnswerFraction": false, "variableReplacements": [], "maxValue": "{a}", "minValue": "{a}", "allowFractions": false, "precisionPartialCredit": 0, "type": "numberentry", "correctAnswerStyle": "plain", "showCorrectAnswer": true, "precisionType": "dp", "showPrecisionHint": true, "precision": "1", "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "precisionMessage": "You have not given your answer to the correct precision.", "scripts": {}}]}], "rulesets": {}, "name": "Arithmetic progressions: simultaneous equations #2", "functions": {}, "extensions": [], "ungrouped_variables": ["n1", "s1", "n2", "s2", "d", "a"], "advice": "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)\\).
\nThe sum of the first \\(\\var{n1}\\) terms of an arithmetic progression is \\(\\var{s1}\\)
\n\\(\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\\)
\n\\(\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}d=\\var{s1}\\) equation (i)
\nThe sum of the first \\(\\var{n2}\\) terms of an arithmetic progression is \\(\\var{s2}\\)
\n\\(\\frac{\\var{n2}}{2}(2a+\\simplify{{n2}-1}d)=\\var{s2}\\)
\n\\(\\var{n2}a+\\simplify{{n2}*({n2}-1)/2}d=\\var{s2}\\) equation (ii)
\nWe can eliminate the \\(a\\) term.
\n\\(\\simplify{{n2}*{n1}}a+\\simplify{{n2}*{n1}*({n1}-1)/2}d=\\simplify{{n2}*{s1}}\\) equation (i) * \\(\\var{n2}\\)
\n\\(\\simplify{{n2}*{n1}}a+\\simplify{{n1}*{n2}*({n2}-1)/2}d=\\simplify{{n1}*{s2}}\\) equation (ii) * \\(\\var{n1}\\)
\nSubtracting gives
\n\\(\\simplify{{n2}*{n1}*({n1}-{n2})/2}d=\\simplify{{n2}*{s1}-{n1}*{s2}}\\)
\n\\(d=\\frac{\\simplify{{n2}*{s1}-{n1}*{s2}}}{\\simplify{{n2}*{n1}*({n1}-{n2})/2}}\\)
\n\\(d=\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))}\\)
\nInserting this value in for \\(d\\) in equation (i) gives
\n\\(\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}(\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))})=\\var{s1}\\)
\n\\(\\var{n1}a+(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})=\\var{s1}\\)
\n\\(\\var{n1}a=\\var{s1}-(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})\\)
\n\\(\\var{n1}a=\\simplify{{s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))}\\)
\n\\(a=\\simplify{({s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2})))/{n1}}\\)
\n\\(a=\\var{a}\\)
\n", "type": "question", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}]}