A thief runs with

Let us assume total time be n minutes

Also, total distance covered by the thief in n minutes =Speed × Time

= 100 × n

= 100n m

And, Total distance covered by policeman = 1^st min. + 2^nd min + …… (n – 1) terms

Here, we have:

a = 100

d = 110 – 100
= 10

‘n’ = n – 1

We know that,

⇒200n = (n-1)[200 + (n-2)10]

⇒200n = (n – 1) [200 + 10n – 20]

⇒10n² + 180n – 10n – 180 – 200n = 0

⇒10n² – 30n – 180 = 0

⇒10 (n² – 3n – 18) = 0

⇒n² – 3n – 18 = 0

⇒n (n – 6) + 3 (n – 6) =0

⇒(n + 3) (n – 6) = 0

∴ n + 3 = 0

n = - 3

Or n – 6 = 0

n = 6

As the value of n i.e., time cannot be negative therefore n = 6

∴ Time taken by the police to catch the thief = n – 1

= 6 – 1

= 5 minutes