Determine 4 distinct numbers $a_{1}, a_{2}, a_{3}$ and $a_{4}$ that are consecutive terms of an arithmetic progression and that the numbers $a_{1}, a_{3}$ and $a_{4}$ form a geometric progression.

## List 8
The 4 terms of an arithmetic progression with common difference $r$ can be written as:

$$
x-2 r, x-r, x, x+r
$$

Therefore, the 3 terms of the geometric progression with common ratio $q$ will be

$$
x-2 r, x, x+r
$$

where

$$
x=(x-2 r) q \text { and } x+r=x q
$$

From this, it follows that:

$$
x=x q-2 r q \Rightarrow x=x+r-2 r q \Rightarrow q=\frac{1}{2}
$$

We obtain that $x+r=\frac{x}{2} \Rightarrow r=-\frac{x}{2}$. Thus, the arithmetic progression is of the form:

$$
2 x, \frac{3 x}{2}, x, \frac{x}{2} \text {. }
$$

Choosing a value for $x$, for example $x=1$, we get 4 numbers forming an arithmetic progression $2,3 / 2,1,1 / 2$ with common difference $-1 / 2$ such that $2,1,1 / 2$ form a geometric progression with common ratio $1 / 2$. Note that this problem has a solution for each choice of $x$, therefore it has an infinite number of solutions.

## List 8