Let a1,a2,...,a100 be non-zero real numbers such that a1+a2+...+a100=0. Then
A
100∑i=1ai2ai>0and100∑i=1ai2−ai<0
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B
100∑i=1ai2ai≥0and100∑i=1ai2−ai≥0
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C
100∑i=1ai2ai≤0and100∑i=1ai2−ai≤0
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D
the sign of 100∑i=1ai2aior100∑i=1ai2−ai depends on the choice of ai's
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Solution
The correct option is A100∑i=1ai2ai>0and100∑i=1ai2−ai<0 For every real numberai,ai2ai>ai&ai2−ai<ai⇒100∑i=1ai2ai>100∑i=1ai&100∑i=1ai2−ai<100∑i=1ai⇒100∑i=1ai2ai>0&100∑i=1ai2−ai<0