A stone falls from rest. The distance covered by the stone in the last second of its motion equals the distance covered by it during the first three seconds of its motion. How long  (in seconds) does the stone take to reach the ground? $\mathrm{Take} \mathrm{g}=10{\mathrm{ms}}^{-2}$

Let n seconds be the time taken by the stone to reach the ground. 

The distance covered in the last second = distance covered in n seconds - distance covered in (n-1) seconds $=\frac{1}{2}{\mathrm{gn}}^2-\frac{1}{2}\mathrm{g}(\mathrm{n}-1{)}^2$.

The distance covered in the first 3 seconds = $\frac{1}{2}\mathrm{g}\times (3{)}^2$.

It is given that  $\frac{1}{2}{\mathrm{gn}}^2-\frac{1}{2}\mathrm{g}(\mathrm{n}-1{)}^2=\frac{1}{2}\mathrm{g}\times (3{)}^2$     or      ${\mathrm{n}}^2-(\mathrm{n}-1{)}^2=9$
Which gives n = 5 s.