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Solving for a geometric series

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\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

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\\(S_{\\simplify{2*{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}=\\var{s_2}\\)

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If we divide one by the other we get:

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\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}}{\\frac{a(1-r^{\\var{n}})}{1-r}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{S_{\\simplify{2*{n}}}}{S_{\\var{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}*\\frac{1-r}{a(1-r^{\\var{n}})}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{1-r^{\\simplify{2*{n}}}}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(\\frac{(1-r^\\var{n})(1+r^{\\var{n}})}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(1+r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)

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\\(r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}-1\\)

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\\(r^{\\var{n}}=\\simplify{{s_2}/{s_1}-1}\\)

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\\(r=\\simplify{(({s_2})/{s_1}-1)^{1/{n}}}\\)

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\\(r=\\simplify{(({s_2}-{s_1})/{s_1})^{1/{n}}}=\\var{r}\\)

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Recall \\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)

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\\(a=\\frac{\\var{s_1}*(1-{r})}{1-r^{\\var{n}}}\\)

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Inserting the value for \\(r\\) in this equation gives

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\\(a=\\frac{\\var{s_1}*(\\simplify{(1-{r})})}{\\simplify{{1-r^{{n}}}}}\\)

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\\(a=\\var{a}\\)

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The sum of the first \\(\\var{n}\\) terms of a geometric series is \\(\\var{s_1}\\) and the sum of the first \\(\\simplify{2*{n}}\\) terms is \\(\\var{s_2}\\).

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

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Calculate the value of the first term.    \\(a\\) = [[1]]

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