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2014-05-04T02:45:36+03:00
A/2+(a+2)/2+(a+4)/2=(a+2)+53
(3a+6)/2-a=55
(3a-2a+6)/2=55
a+6=55·2
a=110-6
a=104
deci nr.pare consecutive sunt 104, 106,108
5 3 5