Answer:
Let total time be n minutes Since policeman runs after 1 minutes so he will catch the thief in \[(n-1)\] minutes, Total distance covered by thief \[=100\text{ }m/minute\times n\] minute \[=(100\text{ }n)\text{ }m\] Now, total distance covered by the policeman \[=(100)m+(100+10)m+(100+10+10)m+.....+(n-1)\] terms i.e., \[100+110+120+....+(n-1)\] terms \[\therefore \,\,{{S}_{n-1}}=\frac{n-1}{2}[2\times 100+(n-2)10]\] \[\Rightarrow \,\frac{n-1}{2}[200+(n-2)10]=100\,n\] \[\Rightarrow \,\,(n-1)(200+10n-20)=200n\] \[\Rightarrow \,\,200n-200+10{{n}^{2}}-10n+20-20n=200n\] \[\Rightarrow \,\,10{{n}^{2}}-30n-180n=0\] \[\Rightarrow \,\,{{n}^{2}}-3n-18=0\] \[\Rightarrow \,\,{{n}^{2}}-(6-3)n-18=0\] \[\Rightarrow \,\,{{n}^{2}}-6n+3n-18=0\] \[\Rightarrow \,\,n(n-6)+3(n-6)=0\] \[\Rightarrow \,\,(n+3)(n-6)=0\] \[\therefore n=6\] or \[n=-3\] (Neglect) Hence, policeman will catch the thief in \[(6-1)\] i.e., 5 minutes.\
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