Assume $a_1 < a_2$ and let $r = \frac{a_2}{a_1}$. Then $r > 1$ and $r$ is a rational number which can be written as $\frac{p}{q}$ with $p, q$ coprime. The sequence is then
$$a_1,\ a_1 r,\ a_1 r^2,\ a_1 r^3,\ a_1 r^4,\ a_1 r^5,\ a_1 r^6.$$
The last term can only be an integer if $a_1$ is divisible by $q^6$. We may try values for $r$ that are equal to $(n+1)/n$ with $n \ge 2$. The first would be $r = 3/2$, but $(3/2)^6 = \frac{729}{64} > 10$ and so $a_1 r^6$ would have more digits than $a_1$. Therefore, $q > 2$.

Next we try $r = 4/3$. Then $a_1$ needs to be divisible by $3^6 = 729$. The smallest value for $a_1$ then is $2 \cdot 3^6 = 1458$. Because $2 \cdot 4^6 = 8192$, we indeed get seven four-digit numbers:

| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|-------------|------|------|------|------|------|------|------|
| $a_n$ | 1458 | 1944 | 2592 | 3456 | 4608 | 6144 | 8192 |

In general, there will be a positive integer $a$ such that $a_1 = a q^6$ and $a_7 = a p^6$. Because $5^6$ is a five-digit number, neither $p$ nor $q$ may exceed 4. It now is easy to see that we found the only possible solution with $r > 1$.