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13问答网 > 设数列an为等比数列且各项均为正数,a1+a2+a3+a4+a5=211⼀27,1⼀a1+1⼀a2+1⼀a3+1⼀a4+1

设数列an为等比数列且各项均为正数,a1+a2+a3+a4+a5=211⼀27,1⼀a1+1⼀a2+1⼀a3+1⼀a4+1

2025-03-16 23:11:14

推荐回答（1个）

回答1：

因为｛an｝是等比数列所以a3的平方等于a1乘以a5 ，也等于a2乘以a41/a1+1/a5=（a1+a5）/(a1*a5) 同理 1/a2+1/a4=(a2+a4) / (a2*a4) 因为 a3的平方等于a1*a5 ，也等于a2*a4 1/a1+1/a2+1/a3+1/a4+1/a5=211/48所以（a1+a5）/(a3的平方) + (a2+a4) / (a3的平方)+a3/ a3的平方=211/48所以（a1+a2+a3+a4+a5）/(a3的平方)=211/48因为 a1+a2+a3+a4+a5=211/27 所以 a3的平方=（211/27 ）/(211/48)=16/9因为｛an｝是等比数列，且各项均为正所以 a3=4/3

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