Let the m-th and n-th terms of a geometric progression be 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

Let the m-th and n-th terms of a geometric progression be 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 Correct Answer -2

Calculation:

Let the first term of GP be a 

According to question

⇒ ar^{m - 1} = 3/4      ----(1)

⇒ ar^{n - 1} = 12      ----(2)

Divide equation 1 by 2

⇒ r^{m - 1 - n + 1} = 1/16

⇒ r^{m - n} = 1/16

⇒ rⁿ - m = 16

So for minimum posssible value we take r = – 4 and n – m = 2

⇒ Minimum posible value of (r + n – m) = – 4 + 2 = – 2

∴ The smallest possible value of r + n - m is – 2