We will prove the statement using induction on $k$.

For $k = 0$ we need to choose four numbers from $\{1, 2, 3, 4, 5\}$ so that among them don't exist three consecutive members of an arithmetic progression. The only such choice is $\{1, 2, 4, 5\}$.

Assume that for some $k \ge 0$ we found a set $S_k$ that contains $4 \cdot 2^k$ positive integers not greater than $5 \cdot 3^k$ so that among them don't exist three consecutive members of an arithmetic progression.

Let $T_k = \{10 \cdot 3^k + s : s \in S_k\}$. Sets $S_k$ and $T_k$ have the same number of elements. We claim that $S_{k+1} = S_k \cup T_k$ satisfies the conditions for $k+1$.

The set $S_{k+1}$ has $2 \cdot (4 \cdot 2^k) = 4 \cdot 2^{k+1}$ elements, and none of them is greater than $10 \cdot 3^k + 5 \cdot 3^k = 5 \cdot 3^{k+1}$. If we take three arbitrary numbers $a, b, c \in S_{k+1}$ such that $a < b < c$, exactly one of the following possibilities is true:
$$1^\circ \quad a, b, c \in S_k$$
$$2^\circ \quad a, b \in S_k, \quad c \in T_k$$
$$3^\circ \quad a \in S_k, \quad b, c \in T_k$$
$$4^\circ \quad a, b, c \in T_k.$$
To prove that the statement holds for $k+1$ it is sufficient to show that in each case $a+c \neq 2b$.

1° $a+c \neq 2b$ because of the induction hypothesis for $S_k$.

$$2^\circ \quad a+c \ge 1 + (10 \cdot 3^k + 1) > 10 \cdot 3^k \ge 2b,$$
$$3^\circ \quad a+c \le 5 \cdot 3^k + 5 \cdot 3^{k+1} = 20 \cdot 3^k < 2b.$$
4° $a - 10 \cdot 3^k, b - 10 \cdot 3^k$ and $c - 10 \cdot 3^k$ are elements of the set $S_k$. Therefore
$$(a - 10 \cdot 3^k) + (c - 10 \cdot 3^k) \neq 2 (b - 10 \cdot 3^k),$$
so $a+c \neq 2b$.