we are given 201 different numbers such that sum of any 100 numbers is less than the sum of remaining 101 numbers. prove that all the numbers are positive .
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1let\ a_1,a_2...a_{201} \ be\ the\ no's\ such\ that\ a_{1}>a_{2}>...>a_{200}>a_{201}\\ \\ given\ a_1+a_2+...a_{100}<a_{101}+...+a_{201}....(i)\\ \\ and\ similarly\ a_{101}+a_{102}+...+a_{200}<a_1+a_2+...+a_{100}+a_{201}\\ \\ adding\ a_{201}\ to\ both\ sides\\ \\ a_{101}+a_{102}+...+a_{200}+a_{201}<a_1+a_2+...+a_{100}+2a_{201} ....(ii)\\ \\ from\ (i)\ &\ (ii)\\ \\ a_1+a_2+...a_{100}<a_1+a_2+...a_{100}+2a_{201}\\ \\ =>a_{201}>0 \\ \\ now, minimum\ of\ all\ 201\ no's\ is\ positive\ hence\ all\ the\ no's\ should\ be\ positive..
