A point traversed half of the distance with a velocity $${v_0}.$$ The half of remaining part of the distance was covered with velocity $${v_1}\& $$  second half of remaining part by $${v_2}$$ velocity. The mean velocity of the point, averaged over the whole time of motion is

Solution :
Let the total distance be $$d.$$ Then for first half distance, time $$ = \frac{d}{{2{v_0}}},$$   next distance. $$ = {v_1}t$$  and last half distance $$ = {v_2}t$$
$$\therefore {v_1}t + {v_2}t = \frac{d}{2};t = \frac{d}{{2\left( {{v_1} + {v_2}} \right)}}$$
Now average speed
$$\eqalign{ & t = \frac{d}{{\frac{d}{{2{v_0}}} + \frac{d}{{2\left( {{v_1} + {v_2}} \right)}} + \frac{d}{{2\left( {{v_1} + {v_2}} \right)}}}} \cr & = \frac{{2{v_0}\left( {{v_1} + {v_2}} \right)}}{{\left( {2{v_0} + {v_1} + {v_2}} \right)}} \cr} $$