Question

已知a,b,c的倒数成等差数列,证明b+c-a/a,a+c-b/b,a+b-c/c也为等差数列？

Answer 1

在abc的倒数同乘a+b+c,然后各减去2,就证完了,7,a，b，c的倒数成等差数列，
∴1/a+1/c=2/b
b=2ac/(a+c)
2(a+c-b)/b=[2(a+c)/b] -2=(a+c)²/ac-2
(b+c-a)/a + (a+b-c)/c
=(b+c)/a + (a+b)/c -2
=[(a+c)b+a²+c²]/ac -2
=[a²+2ac+...,1,所以 2/b=1/a+1/c
所以( b+c-a)/a+(a+b-c)/a-2(a+c-b)/b
=(b+c)/a+(a+b)/c-2(a+c)/b
=b(1/a+1/c)+c/a+a/c-(a+c)(1/a+1/c)
=b*(2/b)+c/a+a/c-1-c/a-a/c-1
=0
所以 是等差数列,1,