A thief runs with a uniform speed of 100 m/min . After 1 minute , a policeman runs after the thief to catch him . He runs with a speed of 100m/ min in the first minute and increases his speed by 10m/min every succeeding minute. After how many minutes will the policeman catch the thief ?

Solution   Let the thief run for n minutes before being caught.

Distance covered by the thief in n minutes=(100n)m

The police man ran for (n-1) minutes to catch the thief.

Distance covered by police man in (n-1) min

= { 100 + 110 + 120 + … to ( n -1 ) terms = S n-1

This is an AP with a = 100 and d = 10

S n -1 = 100 + 110 + 120  + … to ( n -1 ) terms

= n – 1 / 2 × ( 2 × 100 + ( n – 1 – 1 ) × 10 )

= ( n – 1 ) ( 100 + 5n – 10 ) = ( n – 1 ) ( 5n + 90)

= 5n2 + 85n – 90

Clearly , the distance covered by the thief in n min is equal to the distance covered by the police man in ( n – 1 ) min

100n = 5n2 + 85n – 90

5n2 – 15n – 90 = 0

5 ( n2 – 3n – 18)= 0

n2 – 3n – 18 = 0

( n – 6 ) ( n + 3) = 0

n – 6 = 0 or n + 3 = 0

n = 6 or n = -3

So n = 6

Hence , time taken by the policeman to catch the thief = ( n – 1) min = ( 6 – 1 )min

= 5 min