1. Given $a>b>2$, and $a+b$, $a-b$, $a b$, $\frac{b}{a}$ form a geometric sequence in a certain order. Then $a=$ $\qquad$ .
$$
-1.7+5 \sqrt{2} .
$$

Since $a>b>2$, we have
$$
\begin{array}{l}
a b-(a+b)=(a-1)(b-1)-1>0, \\
\frac{b}{a}b^{2}=\frac{b}{a} \cdot a b$, then when the common ratio is greater than 1,
it can only be $\frac{b}{a} 、 a-b 、 a+b 、 a b$ forming a geometric sequence, at this time,
$$
\begin{array}{l}
(a-b)(a+b)=a b \cdot \frac{b}{a} \Rightarrow a=\sqrt{2} b . \\
\text { Hence }(a-b)^{2}=(3-2 \sqrt{2}) b^{2} \\
=\frac{b}{a}(a+b)=\frac{\sqrt{2}+1}{\sqrt{2}} b \\
\Rightarrow b=\frac{7+5 \sqrt{2}}{\sqrt{2}} \\
\Rightarrow a=7+5 \sqrt{2} .
\end{array}
$$