The ratio’s of the distances covered by a freely falling particle released from rest, in the first, second ……. ${\text{n}}^{\text{th}}$  second of its motion.

$\text{u}=0$

$\text{a}=\text{g}$  

Then distances covered in ${\text{n}}^{\text{th}}$ second is ${\text{h}}_{\text{n}}=\text{g}\left(\text{n}-\frac{1}{2}\right)$       

$\text{n}=1{\text{h}}_1=\text{g}(1-\text{1/2)}=\text{g}/2$       

$\text{n}=2{\text{h}}_2=\text{g}(2-\text{1/2)}=3\text{g}/2$

$\text{n}=3{\text{h}}_3=\text{g}(3-\text{1/2)}=5\text{g}/2$  

The ratio of ${\text{h}}_{\text{1}}:{\text{h}}_{\text{2}}:{\text{h}}_{\text{3}}:{\text{h}}_{\text{4}}\mathrm{.......}$  $=1:3:5:7:9:\mathrm{........}$