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Question 70
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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