CAT 2020 Progression Question (Hard)

Question

Let the m-th and n-th terms of a Geometric progression be and 12, respectively, when m n. If the common ratio of the progression is an integer r, then the smallest possible value of r n - m is

Accepted Answer

Step 1 Let the first term be a and common ratio be r an integer Then a r 3 4 1 a r 12 2 Divide 2 by 1 r 12 3 4 16 So r 16 2 Smallest integer r possible is 2 Then n m 4 So r n m 2 4 6 Try r 2 2 16 Only works if n m even 4 valid Now plug r 2 and n m 4 into 1 and 2 Let s compute backward From 1 a 2 3 4

CAT 2020

Quantitative

Progression

Hard

Let the m-th and n-th terms of a Geometric progression be $\frac{3}{4}$ and 12, respectively, when m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is

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