(7) Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and $a_{1}+a_{2}=\frac{3}{4}$, $a_{3}+a_{4}+a_{5}+a_{6}=15$. Then $a_{7}+a_{8}+a_{9}=$ $\qquad$ .
7112 Prompt: Let the common ratio be $q$, then from the given conditions we can get
$$
\begin{array}{c}
a_{1}(1+q)=\frac{3}{4}, \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15,
\end{array}
$$

By comparing these two equations, we get $q^{2}\left(1+q^{2}\right)=20$, thus $q=2, a_{1}=\frac{1}{4}$. Therefore,
$$
a_{7}+a_{8}+a_{9}=a_{1} q^{6}\left(1+q+q^{2}\right)=112 .
$$