$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$     and $$^n{C_{r + 1}} = 126,$$   then $$r$$ is:

Solution :
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84,{\,^n}{C_{r + 1}} = 126$$
KEY CONCEPT : We know that
$$\eqalign{ & \frac{{^n{C_{r - 1}}}}{{^n{C_r}}} = \frac{r}{{n - r + 1}} \cr & \Rightarrow \,\,\frac{{36}}{{84}} = \frac{r}{{n - r + 1}} \cr & \Rightarrow \,\,\frac{r}{{n - r + 1}} = \frac{3}{7} \cr & \Rightarrow \,\,3n - 10r + 3 = 0\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Also }}\frac{{^n{C_r}}}{{^n{C_{r + 1}}}} = \frac{{r + 1}}{{n - r}} = \frac{{84}}{{126}} = \frac{2}{3} \cr & \Rightarrow \,\,2n - 5r - 3 = 0\,\,\,\,\,.....\left( 2 \right) \cr} $$
Solving (1) and (2), we get
$$n = 9$$  and $$r = 3.$$