16. 等比数列 $\left\{a_{n}\right\}$ 的公比为 $q$, 其前 $n$ 项的积为 $T_{n}$, 并且满足条件 $a_{1}>1, a_{99} a_{100}-1>0$, $\frac{a_{99}-1}{a_{100}-1}<0$. 给出下列结论: (1) $0<q<1$; (2) $a_{99} a_{101}-1<0$; (3) $T_{100}$ 的值是 $T_{n}$ 中最大的; (4)使 $T_{n}>1$成立的最大自然数 $n$ 等于 198其中正确的结论是 \$ \qquad \$ . (填写所有正确的序号)

## 答案 (1)(2)(4)
(1)中, $\left\{\begin{array}{l}\left(a_{99}-1\right)\left(a_{100}-1\right)<0 \\ a_{99} a_{100}>1 \\ a_{1}>1\end{array} \Rightarrow\left\{\begin{array}{l}a_{99}>1 \\ 0<a_{100}<1\end{array}\right.\right.$

$\Rightarrow q=\frac{a_{100}}{a_{99}} \in(0,1), \therefore$ (1)正确.

(2)中, $\left\{\begin{array}{l}a_{99} a_{101}=a_{100}^{2} \\ 0<a_{100}<1\end{array} \Rightarrow a_{99} a_{101}<1, \therefore\right.$ (2)正确.

(3)中, $\left\{\begin{array}{l}T_{100}=T_{99} a_{100} \\ 0<a_{100}<1\end{array} \Rightarrow T_{100}<T_{99}, \therefore\right.$ (3)错误.

(4) 中, $T_{198}=a_{1} a_{2} \cdots a_{198}$

$=\left(a_{1} a_{198}\right)\left(a_{2} a_{197}\right) \cdots\left(a_{99} a_{100}\right)$

$=\left(a_{99} a_{100}\right)^{99}>1$,

$T_{199}=a_{1} a_{2} \cdots a_{198} a_{199}=\left(a_{1} a_{199}\right) \cdots\left(a_{99} a_{101}\right) \cdot a_{100}$

$=a 199100<1, \therefore$ (4)正确.