A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of 1m each will then be

The time taken to cover 'n' meters is given by

n = $\frac{1}{2} g{t_n}^2$

$t_n = \sqrt{\frac{2n}{g}}$

Time taken to cover (n+1) meters is given by 

$t_{n+1} = \sqrt{\frac{2(n+1)}{g}}$

Therefore, $t_{n+1}- t_n = \sqrt{\frac{2(n+1)}{g}} - \sqrt{\frac{2n}{g}}$

= $\sqrt{\frac{2}{g}} \left[\sqrt{n+1} - \sqrt{n}\right]$

This gives $the ratio of the difference in the square roots of the integers$$\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})\mathrm{.......}$