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

The correct answer is (A): \(-2\)

\(T_n = 12\)

\(T_m = \frac{3}{4}\)

\(\frac{T_n}{T_m} = \frac{ar^{n-1}}{ar^{m-1}} = \frac{12}{\frac{3}{4}}\)

\(r^{n-m} = 16 = (±2)^4 = (±4)^2\)

To get the minimum value for \(r+n-m, r\) should be minimum.

\(∴ r = -4\)

\(n-m = 2\)

\(∴\)  required answer = \(-2\)

Answer 2

Let "a" be the first phrase in the GP. We can now demonstrate that based on the query.
 \(ar^{m-1}=\frac{3}{4}\\ar^{n-1}=12\)
After dividing both equations, we obtain \(r^{m-1-n+1}=\frac{1}{16}\ i.e, r^{n-m}=16\)
Thus, we determine the lowest number that may be obtained by giving "r" the lowest feasible value of -4 and \(n-m=2.\) 
Thus, the lowest value of \(r+n-m=-4+2=-2\) that can exist.