3. Given that $S_{n}$ is the sum of the first $n$ terms of an arithmetic sequence with a positive common difference $q$. If $\frac{S_{n}+210}{n}$ reaches its minimum value when $n=6$, then the range of values for $q$ is
3. $[10,14]$.

Let $a_{n}=a_{1}+(n-1) q$. Then $S_{n}=n a_{1}+\frac{n(n-1)}{2} q$.
Thus, $\frac{S_{n}+210}{n}=\frac{q}{2} n+\frac{210}{n}+a_{1}-\frac{q}{2}$. Therefore, $\frac{6 q}{2}+\frac{210}{6} \leqslant \min \left\{\frac{5 q}{2}+\frac{210}{5}, \frac{7 q}{2}+\frac{210}{7}\right\}$. This gives $10 \leqslant q \leqslant 14$.