Well, I must say, that's quite the mathematical puzzler! Let's see if I can come up with an answer for you.
If we have a geometric progression between 3 and 9, it means that the ratio between consecutive terms is constant. And if we have an arithmetic progression between the last three terms, it means that the difference between consecutive terms is constant.
So, let's call the two missing numbers a and b. The geometric progression would be 3, a, b, 9, and the arithmetic progression would be a, b, 9.
Given this information, we can set up two equations:
b/a = 9/b (from the geometric progression)
9 - b = b - a (from the arithmetic progression)
Now, let's solve for a and b!
From the first equation, we can cross-multiply to get b^2 = 9a.
Substituting this into the second equation, we get:
9 - b = b - (b^2/9)
Simplifying further, we have:
9 = 2b - (b^2/9)
Rearranging and cross-multiplying, we get:
b^2 - 18b + 81 = 0
This equation does not yield real roots, which means there are no positive numbers that satisfy both the geometric and arithmetic progressions.
Well, I tried my best to find a solution, but it seems like this joke turned out to be a bit of a flop! Maybe math and humor just don't mix well in this case. If you have any other questions or need a good chuckle, feel free to ask!