Clearly the geometric term $r$ has to be greater than 1. Let $1000 \leq a_6 < 10000$. Then we have the following constraints on $r$:
1. $a_1 \geq 1000$, that is, $a_6 \cdot r^{-5} \geq 1000 ~ ~ ~ \implies ~ ~ ~ r \leq \sqrt[5]{\frac{a_6}{1000}}$
2. $a_7 \geq 10000$, which implies $a_6 \cdot r > 10000 ~ ~ ~ \implies ~ ~ ~ r \geq \frac{10000}{a_6}$
3. $a_{10} < 10000$, so $a_6 \cdot r^4 < 100000 ~ ~ ~ \implies ~ ~ ~ r < \sqrt[4]{\frac{100000}{a_6}}$
So, given $1000 \leq a_6 < 10000$, any $r$ satisfying the condition below is a solution to the problem.
$$\frac{10000}{a_6} < r < \min{\left ( \sqrt[5]{\frac{a_6}{1000}}, \sqrt[4]{\frac{100000}{a_6}} \right )}$$
Note that this interval may be empty. I haven't done the calculations but it is obvious that for $a_6$ close to $1000$ there is no solution, and so on. Also the inequalities may be a bit wrong with respect to $\leq$ and $<$ because I'm lazy but I trust it will not be an issue.
Let's try it with $a_6 = 9183$ at random. Then we have:
$$\frac{10000}{a_6} \approx 1.09$$
$$\sqrt[5]{\frac{a_6}{1000}} \approx 1.56$$
$$\sqrt[4]{\frac{100000}{a_6}} \approx 1.82$$
Therefore our condition on $r$ is approximately:
$$1.09 < r < 1.56$$
Arbitrarily, let's pick $r = 1.2$. Then:
$$a_1 = a_6 \cdot r^{-5} \approx 3690$$
And the entire sequence follows, rounded up to integers:
$$a = \{ 3690, 4429, 5314, 6377, 7653, 9183, 11020, 13224, 15868, 19042 \}$$
$$\blacksquare$$
This approach assumes that $a_7$ has 5 digits, this wasn't explicitly specified in the problem and so this method does not capture all solutions (it does capture all solutions under the assumption that $a_7$ has 5 digits, though, as far as I can tell, and should be able to be tweaked to assume that $a_7$ has 4 digits but $a_8$ has 5 and so on).
