Let′sstartcalculationfromthebeginningoffence.Secondpostis10mfarbecausebeginningoffenceandfirstpostareinsimilarposition.Thirdpostis10+10=20mfarfromthebeginning.So,nthpostwillbe10m×(n−1)farenoughfromthebeginning.Let′scalculatewhen10m×(n−1)=2km.10m×(n−1)=2000m⇒n−1=200⇒n=201So,thereare201postsalongthefence.
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