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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}\\).
", "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}\\) equation (i)
\nThe formula for nth term of an arithmetic progression is \\(T_n=a+(n-1)d\\).
\nThe \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\)
\n\\(a+\\simplify{{n2}-1}d=\\var{T}\\) equation (ii)
\nHere we have two simultaneous equations. We can eliminate the \\(a\\) term.
\n\\(\\var{n1}a+\\simplify{({n1}-1)*{n1}/2}d=\\var{s1}\\) equation (i)
\n\\(\\var{n1}a+\\simplify{{n1}*({n2}-1)}d=\\simplify{{n1}*{T}}\\) equation (ii) $\\times$ \\(\\var{n1}\\)
\n\\(\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}d=\\simplify{{s1}-{n1}*{T}}\\)
\n\\(d=\\frac{\\simplify{{s1}-{n1}*{T}}}{\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}}\\)
\n\\(d=\\simplify{({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)
\nUsing this result and equation (ii) we can find the value for \\(a\\)
\n\\(a+\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}=\\var{T}\\)
\n\\(a=\\var{T}-\\simplify{({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)
\n\\(a=\\simplify{{T}-({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)
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