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Solving for a geometric series
"}, "advice": "\\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)
\n\\(S_{\\simplify{2*{n}}}=\\frac{a(1-r^{\\simplify{2*{n}}})}{1-r}=\\var{s_2}\\)
\nIf we divide one by the other we get:
\n\\(\\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}}\\)
\n\\(\\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}}\\)
\n\\(\\frac{1-r^{\\simplify{2*{n}}}}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)
\n\\(\\frac{(1-r^\\var{n})(1+r^{\\var{n}})}{1-r^{\\var{n}}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)
\n\\(1+r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}\\)
\n\\(r^{\\var{n}}=\\frac{\\var{s_2}}{\\var{s_1}}-1\\)
\n\\(r^{\\var{n}}=\\simplify{{s_2}/{s_1}-1}\\)
\n\\(r=\\simplify{(({s_2})/{s_1}-1)^{1/{n}}}\\)
\n\\(r=\\simplify{(({s_2}-{s_1})/{s_1})^{1/{n}}}=\\var{r}\\)
\nRecall \\(S_{\\var{n}}=\\frac{a(1-r^{\\var{n}})}{1-r}=\\var{s_1}\\)
\n\\(a=\\frac{\\var{s_1}*(1-{r})}{1-r^{\\var{n}}}\\)
\nInserting the value for \\(r\\) in this equation gives
\n\\(a=\\frac{\\var{s_1}*(\\simplify{(1-{r})})}{\\simplify{{1-r^{{n}}}}}\\)
\n\\(a=\\var{a}\\)
\n", "tags": [], "functions": {}, "statement": "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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\nCalculate the value of the first term. \\(a\\) = [[1]]
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