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Question:

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

Explanation:

If the first term is a and common ratio is r (an integer)

T_m = ¾ = ar^m-1 and

T_n = 12 = ar^n-1

Now, $\frac{T_n}{T_m}=\frac{12}{3/4}=\frac{a{r}^{n-1}}{a{r}^{m-1}}=r^{n-m}$

∴ r^(n-m) = 16

Since r is an integer, r can be either ± 2, ± 4 or 16.

If r = ± 2, n – m = 4 ⇒ least value of r + n – m = -2 + 4 = 2

If r = ± 4, n – m = 2 ⇒ least value of r + n – m = -4 + 2 = - 2

If r = 16, n – m = 1 ⇒ least value of r + n – m = 16 + 1 = 17

∴ Least possible value of r + n – m = - 2

Hence, option (d).