Question:hard

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

Updated On: Jan 15, 2026
  • -2
  • 2
  • 6
  • -4
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The Correct Option is A

Solution and Explanation

Step 1: Given Data

Consider a geometric progression with first term \( a \) and common ratio \( r \).

The specified terms are:

  • The m-th term is \( ar^{m-1} = \frac{3}{4} \).
  • The n-th term is \( ar^{n-1} = 12 \).

Step 2: Equation Division

Divide the n-th term equation by the m-th term equation:

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

This simplifies to:

\[ r^{n - m} = 16 \]

Step 3: Minimization

We aim to minimize the expression \( r + n - m \), subject to the condition \( r^{n - m} = 16 \). We test possible values:

  • If \( r = -4 \), then \( (-4)^2 = 16 \), which implies \( n - m = 2 \).
  • The expression \( r + n - m \) becomes \( -4 + 2 = -2 \).

✅ Final Answer:

\[ \boxed{-2} \]

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