5. Positive integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ form a strictly increasing geometric sequence, and satisfy $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=211$, then $a_{1}=$ $\qquad$ , common ratio $q=$ $\qquad$ .
Since 211 is a prime number $\Rightarrow q$ is not an integer, and $211=a_{1}+a_{2}+a_{3}+a_{4}+a_{5}>5 a_{1} \Rightarrow a_{1}<42$. Let $q=\frac{m}{n},(m, n)=1$, then $\left.a_{5}=a_{1} q^{4}=\frac{a_{1} m^{4}}{n^{4}} \Rightarrow n^{4} \right\rvert\, a_{1} \Rightarrow n=2, a_{1}=16$. When $a_{1}=16, n=2$, $a_{2} \geqslant 24, a_{3} \geqslant 36, a_{4} \geqslant 54, a_{5} \geqslant 81 \Rightarrow a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \geqslant 211$, equality holds when $q=\frac{3}{2}$, which satisfies the problem.