3. 1. 2 * Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, both consisting of positive numbers, 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 that $2 b_{n}=a_{n}+a_{n+1} \cdots$ (1), $a_{n+1}^{2}=b_{n} \cdot b_{n+1} \cdots$ (2), 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. Since $b_{1}=2, a_{2}=3$, and $a_{2}^{2}=b_{1} b_{2}$, we have $b_{2}=\frac{9}{2}$, thus $\sqrt{b_{n}}=\sqrt{2}+(n-1)\left(\sqrt{\frac{9}{2}}-\sqrt{2}\right)=\frac{\sqrt{2}}{2}(n+1)$. Therefore, $b_{n}=\frac{(n+1)^{2}}{2}, a_{n}=\sqrt{b_{n-1} b_{n}}=\sqrt{\frac{n^{2}}{2} \cdot \frac{(n+1)^{2}}{2}}=\frac{n(n+1)}{2}(n \geqslant 1)$.