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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}\\).

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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}\\)                               equation (i)

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The formula for nth term of an arithmetic progression is \\(T_n=a+(n-1)d\\).

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The \\(\\var{n2}th\\) term of the same series is \\(\\var{T}\\)

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\\(a+\\simplify{{n2}-1}d=\\var{T}\\)                                                   equation (ii)

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Here we have two simultaneous equations. We can eliminate the \\(a\\) term.

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

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\\(\\var{n1}a+\\simplify{{n1}*({n2}-1)}d=\\simplify{{n1}*{T}}\\)                  equation (ii) $\\times$ \\(\\var{n1}\\)

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\\(\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}d=\\simplify{{s1}-{n1}*{T}}\\)

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\\(d=\\frac{\\simplify{{s1}-{n1}*{T}}}{\\simplify{({n1}-1)*{n1}/2-{n1}*({n2}-1)}}\\)

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\\(d=\\simplify{({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)

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Using this result and equation (ii) we can find the value for \\(a\\)

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

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

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\\(a=\\simplify{{T}-({n2}-1)*({s1}-{n1}*{T})/(({n1}-1)*{n1}/2-{n1}*({n2}-1))}\\)

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