CAT 2020 Slot 2 QA Question 7: Let the $m$-th and $n$-th terms of a geometric progression be $\dfrac{3}{4}$ and 12, respectively, where $m

Question

Let the $m$-th and $n$-th terms of a geometric progression be $\dfrac{3}{4}$ and 12, respectively, where $m < n$. If the common ratio of the progression is an integer $r$, then the smallest possible v

Accepted Answer

Let the first term be $a$. Then

$a r^{m-1} = \dfrac{3}{4}$
$a r^{n-1} = 12$

Dividing the second equation by the first,
$r^{n-m} = \dfrac{12}{\dfrac{3}{4}}$
$r^{n-m} = 16$

Since $n - m$ is positive and $r$ is an integer, the possibilities include
$r = 2, n-m = 4$
$r = 4, n-m = 2$
$r = -2, n-m = 4$

Let the mmm-th and nnn-th terms of a geometric progression be 34\dfrac{3}{4}43​ and 12, respectively, where m<nm < nm<n. If the common ratio of the progression is an integer rrr, then the smallest possible value of r+n−mr + n - mr+n−m is

Let the $m$ -th and $n$ -th terms of a geometric progression be $\dfrac{3}{4}$ 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