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![[Post New]](/templates/default/images/icon_minipost_new.gif) 22 Mar 2008 19:32:32 IST
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2) distance to the 'i'th tree is given by
d=10+(i-1)5
now for each plant the gardener travels twice the distance...
so
D=2d=10(2+i-1)=10(i+1)
so total distance=summation (i=1 to i=25)
5i(i+1)+10i
=5i2+15i
5(25)2+15(25)
=3500m
but he does not make last trip from last tree to well=120+10 =130 m
so min dist=3500-130=3370m
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