Question Detail

For three consecutive terms of a GP, the square of the middle term equals the product of the other two: ((2m-1)^2 = m(3m-5)). Expand: (4m^2 -4m +1 = 3m^2 -5m) (\Rightarrow m^2 + m + 1 = 0). Discriminant is negative, so no real m. Possibly the terms are (k-2, k-1, 3k-5) as in a previous question. If it's (m, m-1, 3m-5): ((m-1)^2 = m(3m-5)) (\Rightarrow m^2 -2m+1 = 3m^2-5m) (\Rightarrow 0 = 2m^2 -3m -1) (\Rightarrow m = \frac{3 \pm \sqrt{17}}{4}). Not matching. With (m, 2m-1, 3m-5): ((2m-1)^2 = m(3m-5)) gives (m^2 + m + 1=0), no real. Given the options, the correct pair from a similar problem is 2 and 2/3.