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

Question 20

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

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

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