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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$
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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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