Given that, $\text{m-th},$ and $\text{n-th}$ term of a geometric progression be $\frac{3}{4}$ and $12.$
- The $\text{n-th}$ term of $\text{GP}: \; \text{T}_{n} = ar^{n-1} = 12$
- The $\text{m-th}$ term of $\text{GP}: \; \text{T}_{m} = ar^{m-1} = \frac{3}{4} $
- Where, $a$ is the first term, $r$ is the common ratio.
Now, $ \dfrac{\text{T}_{n}}{\text{T}_{m}} = \dfrac{ar^{n-1}}{ar^{m-1}} = \dfrac{12}{\frac{3}{4}}$
$ \Rightarrow r ^{n – 1 – (m – 1)} = \dfrac{12 \times 4}{3} $
$ \Rightarrow r ^{n – 1 – m + 1} = 16 $
$ \Rightarrow r^{n – m} = 16 $
$ \Rightarrow r^{n – m} = (\pm 4)^{2} = (\pm 2)^{4} $
For the minimum value of $ r + n – m, r $ should be minimum.
So, $ \boxed{r =\; – 4}, n – m = 2 $
Thus, $ r + n – m =\; – 4 + 2 = \; – 2 $
$\therefore$ The smallest possible value of $ r + n – m $ is $\; – 2.$
Correct Answer $: \text{A}$