In the given table, fill in all the cells so that the numbers in each column and each row form a geometric progression.

|  |  | 6 |  |
| :--- | :--- | :--- | :--- |
| 72 |  |  |  |
|  |  |  | 40.5 |
|  | 144 |  |  |
Let the denominator of the progression written in the first row be denoted by $q^{-1}$, then the numbers in the first row are $6 q^{-2}, 6 q^{-1}, 6, 6 q$. From this, the denominators of the progressions in the first, second, and fourth columns are found to be $12 q^{2}, 2 \cdot(3 q)^{1 / 3}$, and $_{3} / 2 \cdot\left(3 q^{-1}\right)^{1 / 2}$, respectively.

Therefore, the second number in the second column is $12 \cdot\left(3 q^{-2}\right)^{1 / 3}$, so the denominator of the progression in the second row is ${ }^{1 / 6} \cdot\left(3 q^{-2}\right)^{1 / 3}$. From this, it follows that the second number in the fourth column is $72 \cdot\left({ }^{1 / 6}\left(3 q^{-2}\right)^{1 / 3}\right)^{3}=q^{-2}$.

On the other hand, it is equal to $6 q \cdot 3 / 2 \cdot\left(3 q^{-1}\right)^{1 / 2}=9 \cdot(3 q)^{1 / 2}$. Therefore, $q^{-4}=35 q$, which means $q=1 / 3$, and the entire table can be uniquely reconstructed.

## Answer

| 54 | 18 | 6 | 2 |
| :---: | :---: | :---: | :---: |
| 72 | 36 | 18 | 9 |
| 36 | 72 | 54 | 40.5 |
| 128 | 144 | 162 | 182.25 |