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Solving for a geometric series

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The second term of a geometric series is given by the formula \\(T_2=ar\\) and the sum to infinity of a geometric series is \\(S_\\infty=\\frac{a}{1-r}\\)

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\\(T_2=ar=\\var{t2}\\)

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\\(a=\\frac{\\var{t2}}{r}\\)

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We can substitute this in for \\(a\\) in the second equation

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\\(S_\\infty=\\frac{a}{1-r}=\\var{s}\\)

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\\(\\frac{\\frac{\\var{t2}}{r}}{1-r}=\\var{s}\\)

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\\(\\frac{\\var{t2}}{r}=\\var{s}(1-{r})\\)

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\\(\\frac{\\var{t2}}{r}=\\var{s}-\\var{s}{r}\\)

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\\(\\var{t2}=\\var{s}r-\\var{s}r^2\\)

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\\(\\var{s}r^2-\\var{s}r+\\var{t2}=0\\) 

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This is a quadratic equation which we can solve by formula.

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\\(r=\\frac{\\var{s}\\pm \\sqrt{(-\\var{s})^2-4*(\\var{s})*(\\var{t2})}}{2*(\\var{s})}\\)

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\\(r=\\frac{\\var{s}+\\sqrt{\\simplify{{s}^2-4*{s}*{t2}}}}{\\simplify{2*{s}}}\\)   or    \\(r=\\frac{\\var{s}-\\sqrt{\\simplify{{s}^2-4*{s}*{t2}}}}{\\simplify{2*{s}}}\\)

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\\(r=\\frac{\\var{s}+\\simplify{({s}^2-4*{s}*{t2})^0.5}}{\\simplify{2*{s}}}\\)    or    \\(r=\\frac{\\var{s}-\\simplify{({s}^2-4*{s}*{t2})^0.5}}{\\simplify{2*{s}}}\\)

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\\(r=\\) {({s}+({s}^2-4*{s}*{t2})^0.5)/(2*{s})}   or    \\(r=\\) {({s}-({s}^2-4*{s}*{t2})^0.5)/(2*{s})}

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\\(a=\\frac{\\var{t2}}{r}\\)

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\\(a=\\) {(2*{s}*{t2})/({s}+({s}^2-4*{s}*{t2})^0.5)}    or    \\(a=\\) {(2*{s}*{t2})/({s}-({s}^2-4*{s}*{t2})^0.5)}

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Calculate the value of the larger common ratio.   \\(r\\) = [[0]]

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Determine the value of the first term of the series corresponding to this common ratio.  \\(a\\) = [[1]]

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Calculate the value of the smaller common ratio.   \\(r\\) = [[2]]

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Determine the value of the first term of the series corresponding to this common ratio.  \\(a\\) = [[3]]

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The second term in a geometric series is \\(\\var{t2}\\) and the sum to infinity of the series is \\(\\var{s}\\).

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There are two possible series that possess these attributes.

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