RagvaloD wrote: $(a_i)$ is arithmetic nonconstant progression, $(b_i)$ is notconstant geometric progression. $a_1=b_1,a_2/b_2=2,a_4/b_4=8$ Find $a_3/b_3$. Writing $a_n=a+(n-1)r$ and $b_n=at^{n-1}$ with $a,r,t\ne 0$ and $t\ne 1$ : $\frac{a_2}{b_2}=2$ implies $t\ne \frac 12$ and $a=\frac r{2t-1}$ $\frac{a_3}{b_3}=8$ implies $t\ne \frac 12$ and $a=\frac {3r}{8t^3-1}$ And so $\frac r{2t-1}=\frac {3r}{8t^3-1}$ which is (since $t\ne\frac 12$) $t=-1$ and $r=-3a$ Then $\boxed{\frac{a_3}{b_3}=-5}$

Sorry, I misread problem. I change conditions in problem.