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Solving for a geometric series

\$$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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