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

Answer: (A)

-2

Answer 2

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 = $-2$