Question 6.

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 $r+n-m = -4+2 = -2$