∴an=8qn-1,
∴bn=log2an=3+(n-1)log2q,
bn+1=3+nlog2q,
∴bn+1-bn=log2q,b1=log28=3,
∴数列{bn}是以3为首项,log2q为公差的等差数列.
由(1)知bn=3+(n-1)log2q,
∵数列{bn}的前n项和中S7最大,且S7≠S8,
∴b7>0,b8<0,
由b7>0,得:3+(7-1)log2q>0,
整理,得2log2q>-1,log2q>-
1 |
2 |
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2 |
由b8<0,得3+(8-1)log2q<0,
整理,得log2q<-
3 |
7 |
1 |
2 |
3 |
7 |
综上,得
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2 |
1 |
2 |
3 |
7 |