1. The arithmetic progression $a_{n}$ is such that its difference is not equal to zero, and $a_{10}, a_{13}$, and $a_{19}$ are consecutive terms of some geometric progression. Find the ratio $a_{12}: a_{18}$.
Solution. Let $a_{n}=a_{1}+d(n-1)$, where $a_{1}$ is the first term of the progression, and $d$ is its common difference.

Then $a_{10}=a_{1}+9 d, a_{13}=a_{1}+12 d, a_{19}=a_{1}+18 d$. By the property of a geometric progression, we have $a_{10} \cdot a_{19}=a_{13}^{2}:\left(a_{1}+9 d\right)\left(a_{1}+18 d\right)=\left(a_{1}+12 d\right)^{2} \rightarrow 18 d^{2}+3 a_{1} d=0 \rightarrow a_{1}=-6 d$. Then

$$
a_{12}: a_{18}=\frac{a_{1}+11 d}{a_{1}+17 d}=\frac{-6 d+11 d}{-6 d+17 d}=\frac{5}{11} .
$$

Answer: $a_{12}: a_{18}=5: 11$.