$\mathrm{A}(\mathrm{a}, \mathrm{b})$ and $\mathrm{B}(0, 0)$ are two fixed points. ${\mathrm{M}}_1$ is the midpoint of $\overline{\mathrm{AB}}, {\mathrm{M}}_2$ is the mid point of $\overline{{\mathrm{AM}}_1}, {\mathrm{M}}_3$ is the midpoint of $\overline{{\mathrm{AM}}_2}$ and so on. Then ${\mathrm{M}}_5$ is

Given points A=(a, b), B=(0, 0)

${\mathrm{M}}_1$ is mid point of $\overline{\mathrm{AB}}=\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)$

${\mathrm{M}}_2$ is mid point of $\overline{{\mathrm{AM}}_1}=\left(\frac{\mathrm{a}+{\displaystyle \frac{\mathrm{a}}{2}}}{2}, \frac{\mathrm{b}+{\displaystyle \frac{\mathrm{b}}{2}}}{2}\right)$

 $=\left(\frac{3\mathrm{a}}{4}, \frac{3\mathrm{b}}{4}\right)$

 ${\mathrm{M}}_3$ is mid point of $\overline{{\mathrm{AM}}_2}=\left(\frac{\mathrm{a}+{\displaystyle \frac{3\mathrm{a}}{4}}}{2}, \frac{\mathrm{b}+{\displaystyle \frac{3\mathrm{b}}{4}}}{2}\right)$

 $=\left(\frac{7\mathrm{a}}{8}, \frac{7\mathrm{b}}{8}\right)$

${\mathrm{M}}_4$ is mid point of $\overline{{\mathrm{AM}}_3}=\left(\frac{\mathrm{a}+{\displaystyle \frac{7\mathrm{a}}{8}}}{2}, \frac{\mathrm{b}+{\displaystyle \frac{7\mathrm{b}}{8}}}{2}\right)$

 $=\left(\frac{15\mathrm{a}}{16}, \frac{15\mathrm{b}}{16}\right)$

and ${\mathrm{M}}_5$ is mid point of $\overline{{\mathrm{AM}}_4}=\left(\frac{\mathrm{a}+{\displaystyle \frac{15\mathrm{a}}{16}}}{2}, \frac{\mathrm{b}+{\displaystyle \frac{15\mathrm{b}}{16}}}{2}\right)$

$=\left(\frac{31\mathrm{a}}{32}, \frac{31\mathrm{b}}{32}\right)$

(or) Apply the formula 

${\mathrm{M}}_{\mathrm{n}}=\left(\frac{2^{\mathrm{n}}-1}{2^{\mathrm{n}}}\mathrm{a}, \frac{2^{\mathrm{n}}-1}{2^{\mathrm{n}}}\mathrm{b}\right)$