If the coefficients of the $\mathrm{r} \mathrm{th}, (\mathrm{r}+1) \mathrm{th}, (\mathrm{r}-2) \mathrm{th}$terms in the expansion of $(1+\mathrm{x}{)}^{14}$are in AP, then the the sum of all possible values of r is ____.

According to the question,
${ }^{14}{\mathrm{C}}_{\mathrm{r}-1}{,}^{14}{\mathrm{C}}_{\mathrm{r}}{,}^{14}{\mathrm{C}}_{\mathrm{r}+1} \mathrm{are} \mathrm{in} A.\mathrm{P}.,\mathrm{so} \left\{\mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}\right\}$
$\Rightarrow 2^{14}{\mathrm{C}}_{\mathrm{r}}{=}^{14}{\mathrm{C}}_{\mathrm{r}-1}{+}^{14}{\mathrm{C}}_{\mathrm{r}+1}$
or   $\frac{2\times 14!}{(14-\mathrm{r})!\mathrm{r}!}=\frac{14!}{(14-\mathrm{r}+1)!(\mathrm{r}-1)!}+\frac{14!}{(14-\mathrm{r}-1)!(\mathrm{r}+1)!}$

or   $\frac{2}{(14-\mathrm{r})(13-\mathrm{r})!\mathrm{r}(\mathrm{r}-1)!}=\frac{1}{(15-\mathrm{r})(14-\mathrm{r})(13-\mathrm{r})!(\mathrm{r}-1)!}+\frac{1}{(13-\mathrm{r})!(\mathrm{r}+1)\mathrm{r}(\mathrm{r}-1)!}$
or   $\frac{2}{(14-\mathrm{r})\mathrm{r}}=\frac{1}{(15-\mathrm{r})(14-\mathrm{r})}+\frac{1}{\mathrm{r}(\mathrm{r}+1)}$
or   $\frac{2}{(14-\mathrm{r})\mathrm{r}}-\frac{1}{\mathrm{r}(\mathrm{r}+1)}=\frac{1}{(15-\mathrm{r})(14-\mathrm{r})}$
or
  $$\begin{gathered} \frac{3\mathrm{r}-12}{\mathrm{r}(\mathrm{r}+1)}=\frac{1}{(15-\mathrm{r})} \\ \mathrm{r}=5\text{ or }9 \end{gathered}$$