题目内容
等比数列{an}的公比为q,其前n项积为Tn,并且满足条件a1>1,a99a100>1,
<0,给出下列结论:①0<q<1;②a99a101-1<0;③T100的值是Tn中最大的;④使Tn>1成立的最大自然数n等于198.其中正确的结论是______.
| a99-1 |
| a100-1 |
①中(a99-1)(a100-1)<0,a1>1,a99a100>1,
?a99>1,0<a100<1?q=
∈(0,1),∴①正确.
②中a99a101=a1002<a100<1?a99a101<1,∴②正确.
③中T100=T99•a100,0<a100<1?T100<T99,∴③错误.
④中T198=a1•a2…a198=(a1•a198)(a2•a197)…(a99•a100)=(a99•a100)99>1,
T199=a1•a2…a199=(a1•a199)(a2•a198)…(a99•a101)a100<1,∴④正确.
答案:①②④
?a99>1,0<a100<1?q=
| a100 |
| a99 |
②中a99a101=a1002<a100<1?a99a101<1,∴②正确.
③中T100=T99•a100,0<a100<1?T100<T99,∴③错误.
④中T198=a1•a2…a198=(a1•a198)(a2•a197)…(a99•a100)=(a99•a100)99>1,
T199=a1•a2…a199=(a1•a199)(a2•a198)…(a99•a101)a100<1,∴④正确.
答案:①②④
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