3. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{7}{6}$, and for any positive integer $n$, the quadratic equation $a_{n} x^{2}-$ $a_{n+1} x+1=0$ has two roots $\alpha_{n}, \beta_{n}$ that satisfy $6 \alpha_{n}-2 \alpha_{n} \beta_{n}+6 \beta_{n}=3$. Then the general term formula for $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$
3. $\frac{1}{2^{n}}+\frac{2}{3}$.

By Vieta's formulas, we have
$$
\alpha_{n}+\beta_{n}=\frac{a_{n+1}}{a_{n}}, \quad \alpha_{n} \beta_{n}=\frac{1}{a_{n}},
$$

thus
$$
3=6\left(\alpha_{n}+\beta_{n}\right)-2 \alpha_{n} \beta_{n}=\frac{6 a_{n+1}-2}{a_{n}},
$$

rearranging, we get
$$
a_{n+1}=\frac{1}{2} a_{n}+\frac{1}{3},
$$

which means
$$
a_{n+1}-\frac{2}{3}=\frac{1}{2}\left(a_{n}-\frac{2}{3}\right).
$$

Since $a_{1}-\frac{2}{3}=\frac{1}{2}$, we have $a_{n}-\frac{2}{3}=\left(\frac{1}{2}\right)^{n}$, i.e., $a_{n}=\frac{1}{2^{n}}+\frac{2}{3}$.