10. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}>0, S_{n}=80$, the largest term among the first $n$ terms is $54, S_{2 n}=$ 6560, the general term formula of this sequence $a_{n}=$ $\qquad$, and $\lim _{n \rightarrow \infty} \frac{a_{n}}{S_{n}}=$ $\qquad$
10. $a_{n}=2 \cdot 3^{n-1}, \frac{2}{3}$.
$S_{n}=\frac{a_{1}}{1-q}\left(1-q^{n}\right), S_{2 n}=\frac{a_{1}}{1-q}\left(1-q^{2 n}\right)$, so $\frac{S_{2 n}}{S_{n}}=1+q^{n}=82$. Thus $q^{n}=81$, and $a_{n}=$ $a_{1} q^{n-1}$ is maximized, $S_{n}=80$, hence $a_{1}=q-1$. Also, $a_{n}=(q-1) q^{n-1}=q^{n}-q^{n-1}=54$, which means $q^{n-1}=27$, therefore $q=3, a_{1}=2$. Thus $a_{n}=a_{1} q^{n-1}=2 \cdot 3^{n-1}, S_{n}=q^{n}-1=3^{n}-1, \lim _{n \rightarrow+\infty} \frac{a_{n}}{S_{n}}=\lim _{n \rightarrow \infty} \frac{2 \cdot 3^{n-1}}{3^{n}-1}$ $=\frac{2}{3}$.