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

We need to work with a geometric progression (GP) and find the smallest possible value of an expression involving the common ratio and term positions.

In a geometric progression, each term is obtained by multiplying the previous term by a fixed number called the common ratio (r).

The general formula for the $k^{th}$ term of a GP is:

$T_k = a \times r^{k-1}$

where $a$ is the first term and $r$ is the common ratio.

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We're given:

$m^{th}$ term = $\tfrac{3}{4}$

$n^{th}$ term = $12$

$m < n$ (so $n - m > 0$)

$r$ is an integer

Using the GP formula:

$a \times r^{m-1} = \tfrac{3}{4}$ ...(1)

$a \times r^{n-1} = 12$ ...(2)

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To eliminate the first term $a$, we divide equation (2) by equation (1):

$\dfrac{a \times r^{n-1}}{a \times r^{m-1}} = \dfrac{12}{\tfrac{3}{4}}$

The $a$ terms cancel out:

$r^{n-m} = 12 \times \tfrac{4}{3} = 16$

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We need to find integer values of $r$ such that $r^{n-m} = 16$.

All ways to express 16 as a power:

$16 = 2^4$ → $r = 2, n-m = 4$

$16 = 4^2$ → $r = 4, n-m = 2$

$16 = 16^1$ → $r = 16, n-m = 1$

$16 = (-2)^4$ → $r = -2, n-m = 4$

$16 = (-4)^2$ → $r = -4, n-m = 2$

For negative values of $r$, we need $n-m$ to be even so that $r^{n-m}$ is positive.

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For each valid pair $(r, n-m)$:

$r = 2, n-m = 4$ → $r + n - m = 2 + 4 = 6$

$r = 4, n-m = 2$ → $r + n - m = 4 + 2 = 6$

$r = 16, n-m = 1$ → $r + n - m = 16 + 1 = 17$

$r = -2, n-m = 4$ → $r + n - m = -2 + 4 = 2$

$r = -4, n-m = 2$ → $r + n - m = -4 + 2 = -2$

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Comparing all possible values: $6, 6, 17, 2, -2$

The smallest possible value is $-2$.

When $r = -4$ and $n-m = 2$, we have $r^{n-m} = (-4)^2 = 16$

Therefore, the smallest possible value of $r + n - m$ is $-2$.