A particle is projected vertically upward with initial velocity 25m/s.during third second of its motion what distance it will travel

If the distance covered in n th second is d, then d= u+(1/2)a(2n+1),where u is initial velocity and a is acceleration. Here a is negative and it`s value is 10 m/s^2. So the answer will be -10 m. The negative sign denotes that the particle will travel 10 m vertically downwards in the third second after attaining its maximum height.