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由題意可得,2a,2b,2c成等差數列∴2b=a+c∴4b2=a2+2ac+c2①∵b2=a2-c2②①②聯立可得,5c2+2ac-3a2=0∵e=ca∴5e2+2e-3=0∵0<e<1∴e=35故答案為:35
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