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}\\)
", "metadata": {"description": "Solving arithmetic progressions using simultaneous equations
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\nCalculate the value of the first term of the series. \\(a\\) = [[1]]
"}], "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) $\\times$ \\(\\var{n2}\\)
\n\\(\\simplify{{n2}*{n1}}a+\\simplify{{n1}*{n2}*({n2}-1)/2}d=\\simplify{{n1}*{s2}}\\) equation (ii) $\\times$ \\(\\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}\\)
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