Given that all terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are positive numbers, and they satisfy: $a_{n} 、 b_{n} 、 a_{n+1}$ form an arithmetic sequence, $b_{n} 、 a_{n+1} 、 b_{n+1}$ form a geometric sequence, and $a_{1}=1, b_{1}=2, a_{2}=3$, find the general terms $a_{n} 、 b_{n}$.
Given:
$$
\begin{array}{l}
2 b_{n}=a_{n}+a_{n+1} \\
a_{n+1}^{2}=b_{n} \cdot b_{n+1}
\end{array}
$$

Since $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are positive sequences, from (2) we get $a_{n+1}=\sqrt{b_{n} b_{n+1}}, a_{n+2}=\sqrt{b_{n+1} \cdot b_{n+2}}$. Substituting into (1) and dividing by $\sqrt{b_{n+1}}$, we get $2 \sqrt{b_{n+1}}=\sqrt{b_{n}}+\sqrt{b_{n+1}}$, which means $\left\{\sqrt{b_{n}}\right\}$ is an arithmetic sequence. Given $b_{1}=2$, $a_{2}=3$, and $a_{2}^{2}=b_{1} b_{2}$, we have $b_{2}=\frac{9}{2}$. Therefore, $\sqrt{b_{n}}=\sqrt{2}+(n-1)\left(\sqrt{\frac{9}{2}}-\sqrt{2}\right)=$ $\frac{\sqrt{2}}{2}(n+1)$. Hence, $b_{n}=\frac{(n+1)^{2}}{2}$, and $a_{n}=\sqrt{b_{n-1} b_{n}}=\sqrt{\frac{n^{2}}{2} \cdot \frac{(n+1)^{2}}{2}}=\frac{n(n+1)}{2}(n \geqslant 1)$.