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Question

Let a1,a2,...,a100 be non-zero real numbers such that a1+a2+...+a100=0. Then

A
100i=1ai2ai>0 and 100i=1ai2ai<0
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B
100i=1ai2ai0 and 100i=1ai2ai0
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C
100i=1ai2ai0 and 100i=1ai2ai0
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D
the sign of 100i=1ai2ai or 100i=1ai2ai depends on the choice of ai's
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Solution

The correct option is A 100i=1ai2ai>0 and 100i=1ai2ai<0
For every real number ai,ai2ai>ai & ai2ai<ai100i=1ai2ai>100i=1ai & 100i=1ai2ai<100i=1ai100i=1ai2ai>0 & 100i=1ai2ai<0

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