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The $n$th term in a geometric progression is given by $a_1 \\times r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio.
\nWe are given that the first term is $\\var{n}$ and the $\\var{dayn}$th term is $\\simplify{{n}*{m}^({dayn}-1)}$
\nHence we know that $r^{\\var{dayn}-1} = \\frac{\\simplify{{n}*{m}^({dayn}-1)}}{\\var{n}} = \\simplify{{m}^({dayn}-1)}$
\nSo we find that $r=\\var{m}$
\n\nIf the pump were to extract air indefinitely, we look to the formula $S_n = \\frac{a_1}{1-r}$ to find $S_n = \\frac{\\var{n}}{1-\\var{m}} = \\simplify{{n}/(1-{m})}$
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