Can any $n$ real numbers be part of a geometric sequence? Ok so, this might be a really silly question, but quite honestly I couldn't find any info about it. So it all begins with a little rather simple question, so I wrote a simple python code, which essentially brute forces step-by-step solutions. The question is Can $x, y, z$ be a part of a geometric sequence? I've noticed a pattern however, for most of the inputs I gave it, it usually took the smallest value from the input, and found a common ratio approaching $1$, ie. $1.001.$ Which effectively approached y and z after multiplying it a lot of times.
Here's an example for input $2,  24$
$24 = 2*12^n$ where $2$ is a, $12$ is r, when we plug it into $ar^n$ we get that $2$ is the first term and $24$ is the second one.
However, aside from that, it spit out many other approximations, such as
$2*1,0001^n$, which also is a very good approximation, and the more time it took to compute the more 0's it added essentially. In here $2$ is the first term, and $24$ is the $24851$st term.
When I let it compute it for an hour with very high approximation it came up with
$2*1,0000001^n$ where $2$ is the first term, and $24$ being $24849068$th term, you get the point.
So now my question is, can any given set of Real positive numbers, with n terms, be expressed as a geometric sequence of $ns*1.0000...1^n$? ns being the smallest nonnegative number?
 A: Suppose $1,2,a$ are terms of a geometric progression. If $r$ is the common ratio of the GP, then $2=r^n$ and $a=r^m$ for some integers $m$ and $n$, so $a^n=2^m$ and $a=2^{m/n}$, a rational power of $2$. Therefore, there are only countably many real numbers that can occur in a geometric progression with $1$ and $2$. In particular, there is no geometric progression containing $1$, $2$, and $3$, since $3^n\ne2^m$.
A: The logarithm of a geometric sequence is an arithmetic sequence. Therefore, for three numbers to be contained in a geometric sequence, the logarithms of those three numbers are contained in an arithmetic sequence. Let the three numbers be $a, b,$ and $c.$ They are all contained in an arithmetic sequence if and only if $\frac{c-a}{b-a}$ is rational. Not every real number is rational, so even three numbers are not guaranteed to all be part of the same geometric sequence. We can easily set $\frac{c-a}{b-a}$ to any real number by making $c$ the real number in question, and making $a$ $0$ and $b$ $1$.
However, we can get arbitrarily good approximations by geometric sequences. To do that, we can just use the method you used and make the common ratio very close to $1.$ In fact we can even get arbitrarily good relative approximations, by log-shifting and using the answer to this question.
