## Task 16/89

We are looking for all arithmetic sequences of the first order where all terms are integers, and the 1st, 3rd, and 4th terms coincide with the correspondingly numbered terms of a geometric sequence.
If there is a sequence $\left\{a_{k}\right\}$ that satisfies the conditions of the problem, then its formation rule is $a_{k}=a_{1}+$ $(k-1) d$, where $a_{1}$ and $d$ are integers; furthermore, there then exists a sequence $\left\{g_{k}\right\}$ with the formation rule $g_{k}=g_{1} \cdot q^{k-1}, g_{1} \neq 0$, and the equations

$$
a_{1}=g_{1} \neq 0 ; \quad a_{3}=a_{1}+2 d=g_{3}=g_{1}^{2} \quad(1) ; \quad a_{4}=a_{1}+3 d=g_{4}=g_{1} q^{3}
$$

From (1) it follows that $d=0.5 a_{1}\left(q^{2}-1\right)$ (3). Substituting this into (2), one obtains, due to $a_{1}=g_{1} \neq 0$, the cubic equation in $q$: $q^{3}-1.5 q^{2}+0.5=0$ with the solutions $q_{1}=q_{2}=1, q_{3}=-0.5$.

From (3) it then follows that $d_{1}=d_{2}=0, d_{3}=-0.375 a_{1}$. The double solution thus yields the trivial case of constant sequences with integer 0. The solution $q=-0.5, d=-0.375 a_{1}$ requires $a_{1} \equiv 0(\bmod 8)$, so $a_{1}=8 a$ with any integer $a \neq 0$, to ensure that $d$ is an integer. Then $d=-3 a$, and the sought (non-trivial) sequences have the formation rule $a_{k}=8 a-3(k-1) a=a(11-3 k)$ with any integer $a \neq 0$.

Addendum: The corresponding geometric sequence then has the formation rule $g_{k}=8 a(-0.5)^{k-1}$ and it is $a_{1}=8 a=g_{1}, a_{3}=2 a=g_{3}, a_{4}=-a=g_{4}$.