A thief away from a police station with a uniform speed 100m/min. After 1 min a policeman runs behind the thief to catch him. He goes at a speed of 100m/min in 1st min and increases the speed 10m/min on each succeeding min. After how many minutes the policeman catches the thief.

Distance traveled by police every one-minute duration from beginning = 100+110+120…+[ 100+(n-1)10]

Above series of distances is in Arithmetic progression.

Sum of the distances of the above Arithmetic progression is same as the distance traveled by thief

(n/2)[ 200+ (n-1)10 ] = (n+1)×100

above expression is simplified as, n^{2} -n -20 =0 or (n-5)(n+4) = 0

hence n=5 is the feasible solution. Thief is caught by police after 6 min