The 4$^\mathrm{th}$ term of GP is 500 and its common ratio is $\frac{1}{m},m\in\mathbb{N}$. Let $\mathrm{S_n}$ denote the sum of the first n terms of this GP. If $\mathrm{S_6 > S_5 + 1}$ and $\mathrm{S_7 < S_6 + \frac{1}{2}}$, then the number of possible values of m is ___________

$T_{4}=500$ <br/><br/> $a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}}$ <br/><br/> Now, <br/><br/> $$ \begin{aligned} & S_{6} > S_{5}+1 \\\\ & \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\ & a r^{5} > 1 \\\\ & \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\ & \Rightarrow \quad m^{2} < 500 \\\\ & \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\ & \quad S_{7} < S_{6}+\frac{1}{2} \\\\ & \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\ & \quad a r^{6}<\frac{1}{2} \end{aligned} $$ <br/><br/> $\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$ <br/><br/> $\frac{1}{m^{3}}<\frac{1}{1000}$ <br/><br/> $\Rightarrow m \in(10, \infty)$ <br/><br/> Possible values of $m$ is $\{11,12,....22 \}$ <br/><br/> $\because m \in N$ <br/><br/> Total 12 values