14. 等比数列 $\left\{a_{n}\right\}$ 同时满足下列三个条件:
(1) $a_{1}+a_{6}=11$
(2) $a_{3} \cdot a_{4}=\frac{32}{9}$
(3)三个数 $\frac{2}{3} a_{2}, a_{3}^{2}, a_{4}+\frac{4}{a}$ 依次成等差数列, 试求数列 $\left\{a_{n}\right\}$的通项公式.
由等比数列的性质知 $a_{1} a_{6}=a_{3} a_{4}=\frac{32}{9}$

$\therefore\left\{\begin{array}{l}a_{1}+a_{6}=11 \\ a_{1} \cdot a_{6}=\frac{32}{9}\end{array}\right.$ 解得 $\left\{\begin{array}{l}a_{1}=\frac{1}{3} \\ a_{6}=\frac{32}{3}\end{array}\right.$ 求 $\left\{\begin{array}{l}a_{1}=\frac{32}{3} \\ a_{6}=\frac{1}{3}\end{array}\right.$

当 $\left\{\begin{array}{l}a_{1}=\frac{1}{3} \\ a_{6}=\frac{32}{3}\end{array}\right.$ 时 $q=2$

$\therefore a_{n}=\frac{1}{3} \cdot 2^{n^{-}}$

$\frac{2}{3} a_{2}+a_{4}+\frac{4}{9}=\frac{32}{9}, \quad 2 a_{3}^{3}=\frac{32}{9}$

$\therefore \frac{2}{3} a_{2}, a_{3}^{2}, a_{4}+\frac{4}{9}$ 成等差数列,
$\therefore a_{n}=\frac{1}{3} \cdot 2^{n^{-1}}$

当 $\left\{\begin{array}{l}a_{1}=\frac{32}{3} \\ a_{6}=\frac{1}{3}\end{array} \quad\right.$ 时 $q=\frac{1}{2}, a_{n}=\frac{1}{3} \cdot 2^{6^{-n}}$

$\frac{2}{3} a_{2}+a_{4}+\frac{4}{9} \neq 2 a_{3}^{2}$,

$\therefore$ 不符合题意,

$\therefore$ 通项公式 $a_{n}=\frac{1}{3} \cdot 2^{n^{-1}}$.

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