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Let the $\text{m-th}$ and $\text{n-th}$ terms  of a geometric progression be $\frac{3}{4}$ and $12,$ respectively, where $\text{m < n}.$ If the common ratio of the progression is an integer $\textsf{r},$ then the smallest possible value of $\textsf{r+n-m}$ is

1. $- 2$
2. $2$
3. $6$
4. $– 4$

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 $: \text{A}$

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