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An insect starts from a point and travels in a straight path 1 mm in the first second and half of the distance covered in the previous second in the succeeding second. In how much time would it reach a point 3 mm away from its starting point. |
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Answer» Distance covered by the insect in the 1st second =1 mm Distance coverd by it in the 2nd second `=1xx(1)/(2)=(1)/(2) " mm "` Distance covered by it in the 3rd second `=(1)/(2)xx(1)/(2)=(1)/(4) " mm "` `" "vdots " "vdots " "vdots " "vdots` The distance covered by the insect in 1st second, 2nd second, 3rd second,`"...."` are respectively `1,(1)/(2),(1)/(4),"....",` which are in GP with `a=1,r=(1)/(2)`. Let time taken by the insect in covering 3 mm be n seconds. `therefore 1+(1)/(2)+(1)/(4)+"...."+n " terms "=3` `implies (1*[1-((1)/(2))^(n)])/(1-(1)/(2))=3` `implies 1-((1)/(2))^(n)=(3)/(2)` `implies ((1)/(2))^(n)=-(1)/(2)` `implies 2^(n)=-2` which is impossible because `2^(n)gt0` `therefore` Our supposition is wrong. `therefore` There is no `n in N`, for which the insect could never 3 mm in n seconds. Hence, it will never to able to cover 3 mm. |
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