Let the m-th and n-thterms of a Geometric progression be 34 and 12

Let the two terms be a1 and a2.
As the terms are in GP with common factor r and say first term as a, a1= ar^(m-1)=3/4 and a2= ar^(n-1)=12
Dividing these two numbers, we get a2/a1= r^(n-m)=16.

Now r can be 2,-2, 4 or -4.
With r=-4, n-m=2. Therefore, r+n-m=-4+2=-2

Given: Let the m-th and n-thterms of a Geometric progression be 3/4 and 12, respectively, when m < n.
Asked: If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is

The m-th and n-thterms of a Geometric progression be 3/4 and 12
ar^{m-1} = 3/4 (1)
ar^{n-1} = 12 (2)
where a is the first term of the sequence and common ratio is r.

(2)/(1)
r^{n-m} = 12*4/3 = 16 = (-2)^4 = 2^4 = (-4)^2 = 4^2 = 16^1

If r=2; n-m = 4; r+n-m = 6
If r=4; n-m = 2; r+n-m = 6
If r=16; n-m = 1; r+n-m = 17
If r=-2; n-m = 4; r+n-m = 2
If r=-4; n-m = 2; r+n-m = -2

Smallest value of r+n-m = -2; If we are taking integral values

IMO B
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We are given that \(\frac{3}{4} * r^a = 12\)
\(r^a = 16 = 2^4 = (-2)^4 = 4^2 = (-4)^2\)
So r could be 2 or -2 in which case (n - m) would be 4 (since r multiplies 4 times with 3/4 to give 12, the two terms must be 4 apart)
Or r could be 4 or -4 in which case (n - m) is 2.
r cannot be 16 since were are looking for the smallest value.

Since we need the smallest value of r + (n-m), it must be -4 + 2 = -2

Answer (B)
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Right substitution of values should take us through
let r=-4 , n=3 and m=1

=>-4+3 -1
=>-2

Therefore IMO B