Question

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

• A. -2
• B. 2
• C. 6
• D. -4

Answer 1

The Correct Option is A

Step 1: Given 

Let the first term of a geometric progression be \( a \) and the common ratio be \( r \).

The following terms of the GP are given:

• \( ar^{m-1} = \frac{3}{4} \) — the m-th term
• \( ar^{n-1} = 12 \) — the n-th term

Step 2: Divide the Two Equations

\[ \frac{ar^{n-1}}{ar^{m-1}} = \frac{12}{\frac{3}{4}} = 16 \Rightarrow r^{n - m} = 16 \]

Step 3: Choose Minimum Value

To minimize \( r + n - m \), we explore different values of \( r \) and \( n - m \) such that: \[ r^{n - m} = 16 \] 
Try values:

• \( r = -4 \Rightarrow (-4)^2 = 16 \Rightarrow n - m = 2 \)
• \( \Rightarrow r + n - m = -4 + 2 = -2 \)

 

✅ Final Answer:

\[ \boxed{-2} \]