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CAT 2020 Slot 2 QA Question & Solution

AlgebraHard

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

Options

6
2
-4
-2

Solution

Let the first term of the GP be "a" . Now from the question we can show that

$ar^{m-1}=\frac{3}{4}$    $ar^{n-1}=12$

Dividing both the equations we get $r^{m-1-n+1}=\frac{1}{16}\ or\ r^{m-n}=16^{-1\ }or\ r^{n-m}=16$

So for the minimum possible value we take Now give minimum possible value to "r" i.e -4 and n-m=2

Hence minimum possible value of r+n-m=-4+2=-2