1 |
a |
1 |
b |
1 |
c |
則
1 |
b |
1 |
a |
1 |
c |
1 |
b |
a?b |
ab |
b?c |
cb |
a?b |
a |
b?c |
c |
又∵a,b,c成等差數(shù)列,且公差不為零,
∴a-b=b-c≠0.由以上兩式,可知
1 |
a |
1 |
c |
兩邊都乘以ac,得a=c.
這與已知數(shù)列a,b,c的公差不為零,a≠c相矛盾,
所以數(shù)列
1 |
a |
1 |
b |
1 |
c |
1 |
a |
1 |
b |
1 |
c |
1 |
a |
1 |
b |
1 |
c |
1 |
b |
1 |
a |
1 |
c |
1 |
b |
a?b |
ab |
b?c |
cb |
a?b |
a |
b?c |
c |
1 |
a |
1 |
c |
1 |
a |
1 |
b |
1 |
c |