Find a geometric progression with members ${1\over 2}$, ${2\over 3}$, and ${3\over 4}$. The question relates to music theory and I have already considered the geometric progression where $a_0=1$ and $\lambda = \sqrt[12]{1\over 2}$.
Using fraction approximations to $\lambda = \sqrt[12]{1\over 2} = 0.9438743126816935$ after forming the corresponding continued fraction, I considered also the fractions ${17 \over 18}$ and ${84 \over 89}$ as possible candidates for such a progression.
In this sense, I am not expecting an exact solution, but can we do better than that avoiding degenarate solutions? Are there different ways of approaching the problem?
 A: It looks like you are looking for a size of semitones that would give the closest match to a just intonation interval.
It's very likely that the sequence wouldn't be geometric. Musicians change the size of semitones all the time to get just intonation. 
Even if there's such a sequence, it will not be practically useful since the vertical alignment of the music will be significantly disturbed as the just intonation tunings of say, C and E, and C and G, will result in an ugly crash between E and G. And that's just the simplest major triad. We will end up with more complications with more complex chords like suspendeds, sevenths and ninths that are present in today's music.
You may want to look into using Cents instead of Hertz to form a sequence or look at other types of sequence that will give you nicer results. There's a nice recurrence relations that you can form if you look at the Wikipedia article. The catch is that it will implies a pattern of varying semitones size instead of a constant semitones you seem to be looking for.
Please update me which pattern you would like to use if you are changing approach! I used to do a project on a similar issue a few years ago too!
A: I do not know if this is what you expect.
If you want to rationalize $\lambda=\sqrt[12]{1\over 2}$ with the smallest denominator that makes the number exact within $10^{-n}$, you would get
$$\left(
\begin{array}{cc}
n & \lambda \\
 2 & \frac{15}{16} \\
 3 & \frac{17}{18} \\
 4 & \frac{84}{89} \\
 5 & \frac{185}{196} \\
 6 & \frac{1194}{1265}
\end{array}
\right)$$
