// Numbas version: exam_results_page_options
{"name": "Arithmetic Series 4", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"parts": [{"correctAnswerFraction": false, "maxValue": "{t1}", "showCorrectAnswer": true, "marks": 1, "integerAnswer": true, "prompt": "<p>Find the first&nbsp;term</p>", "integerPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "scripts": {}, "showPrecisionHint": false, "variableReplacements": [], "type": "numberentry", "allowFractions": false, "minValue": "{t1}"}, {"correctAnswerFraction": false, "maxValue": "{tans}", "showCorrectAnswer": true, "marks": 1, "integerAnswer": true, "prompt": "<p>Find the sum of the first $\\var{tsum}$ terms</p>", "integerPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "scripts": {}, "showPrecisionHint": false, "variableReplacements": [], "type": "numberentry", "allowFractions": false, "minValue": "{tans}"}, {"correctAnswerFraction": false, "maxValue": "ceil((-(2*{t1}-{d})+sqrt((2*{t1}-{d})^2+8000*{d}))/(2*{d}))", "showCorrectAnswer": true, "marks": 1, "integerAnswer": true, "prompt": "<p>Determine the least value of <em>n</em> for which the sum of the first <em>n</em> terms of the series exceeds 1000.</p>", "integerPartialCredit": 0, "variableReplacementStrategy": "originalfirst", "scripts": {}, "showPrecisionHint": false, "variableReplacements": [], "type": "numberentry", "allowFractions": false, "minValue": "ceil((-(2*{t1}-{d})+sqrt((2*{t1}-{d})^2+8000*{d}))/(2*{d}))"}], "name": "Arithmetic Series 4", "question_groups": [{"pickQuestions": 0, "name": "", "pickingStrategy": "all-ordered", "questions": []}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": ""}, "statement": "<p>The $\\var{n}$<sup>th&nbsp;</sup>term of an arithmetic series is $\\simplify{{t1}+({n}-1)*{d}}$</p>\n<p>The common difference is $\\var{d}$</p>", "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["t1", "d", "t2", "t3", "t4", "tr", "tsum", "tans", "n", "a", "b", "bsq", "dscrm", "solvet1", "solvet2", "nc"], "rulesets": {}, "functions": {}, "variables": {"dscrm": {"templateType": "anything", "name": "dscrm", "description": "", "group": "Ungrouped variables", "definition": "sqrt(bsq-(4*a*-1000))"}, "t2": {"templateType": "anything", "name": "t2", "description": "", "group": "Ungrouped variables", "definition": "t1+d"}, "n": {"templateType": "anything", "name": "n", "description": "", "group": "Ungrouped variables", "definition": "random(4..20)"}, "nc": {"templateType": "anything", "name": "nc", "description": "", "group": "Ungrouped variables", "definition": "precround(solvet1+solvet2,2)"}, "d": {"templateType": "anything", "name": "d", "description": "", "group": "Ungrouped variables", "definition": "random(2..9 except t1 )"}, "solvet2": {"templateType": "anything", "name": "solvet2", "description": "", "group": "Ungrouped variables", "definition": "dscrm/(2*a)"}, "solvet1": {"templateType": "anything", "name": "solvet1", "description": "", "group": "Ungrouped variables", "definition": "(-b/(2*a))"}, "t4": {"templateType": "anything", "name": "t4", "description": "", "group": "Ungrouped variables", "definition": "t1+3d"}, "b": {"templateType": "anything", "name": "b", "description": "", "group": "Ungrouped variables", "definition": "(2*t1-d)/2"}, "tsum": {"templateType": "anything", "name": "tsum", "description": "", "group": "Ungrouped variables", "definition": "random(11..29)"}, "bsq": {"templateType": "anything", "name": "bsq", "description": "", "group": "Ungrouped variables", "definition": "b^2"}, "tans": {"templateType": "anything", "name": "tans", "description": "", "group": "Ungrouped variables", "definition": "(tsum)/2*(2*t1+(tsum-1)*d)"}, "tr": {"templateType": "anything", "name": "tr", "description": "", "group": "Ungrouped variables", "definition": "random(6..10)"}, "t3": {"templateType": "anything", "name": "t3", "description": "", "group": "Ungrouped variables", "definition": "t1+2d"}, "a": {"templateType": "anything", "name": "a", "description": "", "group": "Ungrouped variables", "definition": "d/2"}, "t1": {"templateType": "anything", "name": "t1", "description": "", "group": "Ungrouped variables", "definition": "random(2..9)"}}, "preamble": {"js": "", "css": ""}, "tags": [], "variable_groups": [], "type": "question", "advice": "<p>Part a)</p>\n<p>To find the first<sup>&nbsp;</sup>term $a_1$, we need to first start with the formula for the $n^{th}$ term:</p>\n<p>$a_n=a_1+d(n-1)$ $_{..(i)}$&nbsp; &nbsp; which is rearranged in terms of $a_1$: &nbsp; &nbsp;$a_1=a_n-d(n-1)$ $_{..(ii)}$</p>\n<p>We then identify the variables which we know, namely: $n=\\var{n}$, &nbsp; &nbsp;$a_n=\\simplify{{t1}+({n}-1)*{d}}$, &nbsp; &nbsp;$d=\\var{d}$</p>\n<p>These values can then be subsituted into $_{(ii)}$ and solved for $a_1$:</p>\n<p>$a_1=\\simplify{{t1}+({n}-1)*{d}}-\\var{d}(\\var{n}-1)$</p>\n<p>Part b)</p>\n<p>To find the sum of the first $\\var{tsum}$ terms we take the formula for the sum of an arithmetic series:</p>\n<p>$S_n=\\frac{n}{2}(a_1+a_n)$&nbsp;$_{..(iii)}$</p>\n<p>While we don't yet know the value of $a_n$, it is obtainable with formula&nbsp;$_{..(i)}$. It is common to substitute this formula into the formula for the sum as follows:</p>\n<p>$S_n=\\frac{n}{2}(a_1+a_1+d(n-1))$</p>\n<p>$\\therefore\\;\\;\\;S_n=\\frac{n}{2}(2a_1+d(n-1))$ &nbsp; &nbsp;$_{..(iv)}$</p>\n<p>We identify the variables which we know, namely: $n=\\var{tsum}$, &nbsp; &nbsp;$a_1=\\var{t1}$, &nbsp; &nbsp;$d=\\var{d}$</p>\n<p>They are substituted into&nbsp;$_{..(iv)}$ and solved for $S_n$:</p>\n<p>$S_\\var{tsum}=\\frac{\\var{tsum}}{2}(2\\times\\var{t1}+\\var{d}(\\var{tsum}-1))$</p>\n<p>Part c)</p>\n<p>To&nbsp;determine the least value of&nbsp;$n$&nbsp;for which the sum of the first $n$&nbsp;terms of the series exceeds $1000$, the sum formula&nbsp;$_{..(iv)}$ is set up with the lower limit value of $1000$ substituted in:</p>\n<p>$1000=\\frac{n}{2}(2\\times\\var{t1}+\\var{d}(n-1))$</p>\n<p>It is then expanded algebraically to give the following quadratic:</p>\n<p>$\\simplify{{d}/2}n^2+\\simplify{{2*{t1}-{d}}/2}n-1000=0$</p>\n<p>$n$ can then be found using the quadratic formula:</p>\n<p>$n=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}$ &nbsp; &nbsp;where $a=\\simplify{{d}/2}$, &nbsp; &nbsp;$b=\\simplify{{2*{t1}-{d}}/2}$ &nbsp; and $c=-1000$</p>\n<p>Thus,</p>\n<p>$n=\\frac{-\\var{b}\\pm\\sqrt{(\\var{b})^2-\\simplify{{-1000*4*{a}}}}}{\\simplify{2*{a}}}$</p>\n<p>$\\therefore\\;\\;\\;n=-\\left(\\simplify{{2*{t1}-{d}}/{2*2*{a}}}\\right)\\pm\\left(\\frac{\\sqrt{\\simplify{{bsq}-{-1000*4*{a}}}}}{\\simplify{2*{a}}}\\right)$</p>\n<p>In this context, we are talking of positive $n^{th}$ terms, so we disregard the negative square root.</p>\n<p>The result is then computed to be: $\\var{nc}$</p>\n<p>If this is the value of $n$ which produces $1000$, it follows that the next whole number will be the least value of $n$ for which the sum of the first $n$ terms of the series exceeds $1000$.</p>", "extensions": [], "showQuestionGroupNames": false, "contributors": [{"name": "Jinhua Mathias", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/514/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Jinhua Mathias", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/514/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}