The ratio of distance travelled by a freely fallIing  body in 1st sec, 2nd sec, 3rd sec……..

The distance travelled by freely falling body, initial velocity=U=0; $S=\frac{1}{2}g{t}^2$ ;  acceleration due to gravity=g;time elapsed=t;displacement=S

 Distance in 1st sec is $=\frac{g}{2}$
 Distance in 2nd sec is $\frac{4g}{2}-\frac{g}{2}=\frac{3g}{2}$ 
 Distance in 3rd sec is  $\frac{9g}{2}-\frac{4g}{2}=\frac{5g}{2}$
 The ratio of distance is  $\frac{g}{2}:\frac{3g}{2}:\frac{5g}{2}=1:3:5:\mathrm{.....}$