// Numbas version: exam_results_page_options {"name": "Arithmetic progressions: simultaneous equations #2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "

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}\$$

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Solving arithmetic progressions using simultaneous equations

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Calculate the value of the common difference.   \$$d\$$ = [[0]]

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

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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)\$$.

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The sum of the first \$$\\var{n1}\$$ terms of an arithmetic progression is \$$\\var{s1}\$$

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\$$\\frac{\\var{n1}}{2}(2a+\\simplify{{n1}-1}d)=\\var{s1}\$$

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\$$\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}d=\\var{s1}\$$                                       equation (i)

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The sum of the first \$$\\var{n2}\$$ terms of an arithmetic progression is \$$\\var{s2}\$$

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\$$\\frac{\\var{n2}}{2}(2a+\\simplify{{n2}-1}d)=\\var{s2}\$$

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\$$\\var{n2}a+\\simplify{{n2}*({n2}-1)/2}d=\\var{s2}\$$                                      equation (ii)

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We can eliminate the \$$a\$$ term.

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\$$\\simplify{{n2}*{n1}}a+\\simplify{{n2}*{n1}*({n1}-1)/2}d=\\simplify{{n2}*{s1}}\$$                               equation (i) * \$$\\var{n2}\$$

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\$$\\simplify{{n2}*{n1}}a+\\simplify{{n1}*{n2}*({n2}-1)/2}d=\\simplify{{n1}*{s2}}\$$                               equation (ii) * \$$\\var{n1}\$$

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Subtracting gives

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\$$\\simplify{{n2}*{n1}*({n1}-{n2})/2}d=\\simplify{{n2}*{s1}-{n1}*{s2}}\$$

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\$$d=\\frac{\\simplify{{n2}*{s1}-{n1}*{s2}}}{\\simplify{{n2}*{n1}*({n1}-{n2})/2}}\$$

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\$$d=\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))}\$$

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Inserting this value in for \$$d\$$ in equation (i) gives

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\$$\\var{n1}a+\\simplify{{n1}*({n1}-1)/2}(\\simplify{2*({n2}*{s1}-{n1}*{s2})/({n2}*{n1}*({n1}-{n2}))})=\\var{s1}\$$

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\$$\\var{n1}a+(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})=\\var{s1}\$$

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\$$\\var{n1}a=\\var{s1}-(\\simplify{({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))})\$$

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\$$\\var{n1}a=\\simplify{{s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2}))}\$$

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\$$a=\\simplify{({s1}-({n1}-1)*({n2}*{s1}-{n1}*{s2})/({n2}*({n1}-{n2})))/{n1}}\$$

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

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