Let the m-th and n-th terms of a geometric progression be $\frac{3}{4}$ 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

The correct answer is −2.

Let the first term of the geometric progression be a and the common ratio be integer r.

We are given:

• The m-th term: $a\cdot r^{m-1}=\frac{3}{4}$

• The n-th term: $a\cdot r^{n-1}=12$

Dividing the n-th term by the m-th term to eliminate a:

$\frac{a\cdot r^{n-1}}{a\cdot r^{m-1}}=\frac{12}{\frac{3}{4}}=12\times \frac{4}{3}=16$

This simplifies to the key equation:

$r^{n-m}=16$

Since r must be a non-zero integer and n > m (so n − m is a positive integer), we find all integer solutions to $r^{n-m}=16$:

| r | n − m | Check | r + n − m |

• $r=2$, $n-m=4$   →  $2^4=16$ ✓   →  $2+4=6$

• $r=\mathrm{-2}$, $n-m=4$   →  ${\left(\mathrm{-2}\right)}^4=16$ ✓   →  $\mathrm{-2}+4=2$

• $r=4$, $n-m=2$   →  $4^2=16$ ✓   →  $4+2=6$

• $r=\mathrm{-4}$, $n-m=2$   →  ${\left(\mathrm{-4}\right)}^2=16$ ✓   →  $\mathrm{-4}+2=\mathrm{-2}$

• $r=16$, $n-m=1$   →  ${16}^1=16$ ✓   →  $16+1=17$

(Note: $r=\mathrm{-16}$ with $n-m=1$ gives $\left(\mathrm{-16}\right){}^1=\mathrm{-16}\ne 16$, so it is invalid.)

Verification for the minimum case (r = −4, n − m = 2):

Take m = 1, n = 3. Then the first term is:

$a=\frac{3}{4}$

Check the n-th (3rd) term:

$a\cdot r^2=\frac{3}{4}\cdot {\left(\mathrm{-4}\right)}^2=\frac{3}{4}\cdot 16=12$  ✓

Comparing all valid values of r + n − m: {6, 2, 6, −2, 17}, the smallest possible value is:

$r+n-m=\mathrm{-4}+2=\mathrm{<strong>-2</strong>}$