$\underline{\text { Gorayzin }}$ D. $\underline{B}$.

Given two non-constant sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$, one of which is arithmetic and the other is geometric.

It is known that $a_{1}=b_{1}, a_{2}: b_{2}=2$ and

$a_{4}: b_{4}=8$. What can the ratio $a_{3}: b_{3}$ be equal to?
Let $a_{1}=b_{1}=a \neq 0$, the common difference of the arithmetic progression be $d \neq 0$, and the common ratio of the geometric progression be $-q \neq 0$.

1. There are two cases.
1) $\left(a_{n}\right)$ is an arithmetic progression, and $\left(b_{n}\right)$ is a geometric progression. Then, by the condition, $a+d=2 a q, a+3 d=8 a q^{3}$, or $d=a(2 q-1)$

$3 d=a\left(8 q^{3}-1\right)=d\left(4 q^{2}+2 q+1\right), 2 q^{2}+q-1=0$, from which $q=1 / 2$ or $q=-1$.

If $q=1 / 2$, then $d=a(2 q-1)=0$, which contradicts the condition.

If $q=-1$, then $d=-3 a$ and $a_{3} / b_{3}={ }^{a+2 d} / a q^{2}=-5$.

2) $\left(a_{n}\right)$ is a geometric progression, and $\left(b_{n}\right)$ is an arithmetic progression. Then $2(a+d)=a q, 8(a+3 d)=a q^{3}$, so $2 d=a(q-2), 24 d=a\left(q^{3}-8\right)=2 d\left(q^{2}+2 q+4\right), q^{2}+2 q-8=0$, from which $q=2$ or $q=-4$. In the first case, again $d=0$, and in the second case $q=-4, d=-3 a$ and $a_{3} / b_{3}=a q^{2} / a+2 q=-16 / 5$.

## Answer

-5 or $-3.2$.