1. **State the problem:** We are given three terms of a geometric progression (GP): $x-2$, $x-1$, and $3x-5$. We need to find the value of $x$. 2. **Recall the property of a GP:** In a geometric progression, the ratio between consecutive terms is constant. So, the ratio of the second term to the first term equals the ratio of the third term to the second term: $$\frac{x-1}{x-2} = \frac{3x-5}{x-1}$$ 3. **Set up the equation and cross-multiply:** $$(x-1)^2 = (x-2)(3x-5)$$ 4. **Expand both sides:** Left side: $$(x-1)^2 = x^2 - 2x + 1$$ Right side: $$(x-2)(3x-5) = 3x^2 - 5x - 6x + 10 = 3x^2 - 11x + 10$$ 5. **Equate and simplify:** $$x^2 - 2x + 1 = 3x^2 - 11x + 10$$ Bring all terms to one side: $$0 = 3x^2 - 11x + 10 - x^2 + 2x - 1$$ $$0 = 2x^2 - 9x + 9$$ 6. **Solve the quadratic equation:** $$2x^2 - 9x + 9 = 0$$ Use the quadratic formula: $$x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 9}}{2 \cdot 2} = \frac{9 \pm \sqrt{81 - 72}}{4} = \frac{9 \pm \sqrt{9}}{4} = \frac{9 \pm 3}{4}$$ 7. **Find the two possible values:** - $$x = \frac{9 + 3}{4} = \frac{12}{4} = 3$$ - $$x = \frac{9 - 3}{4} = \frac{6}{4} = 1.5$$ 8. **Check for validity:** Check if each $x$ value makes all terms valid and the ratio consistent. - For $x=3$: Terms: $3 - 2 = 1$, $3 - 1 = 2$, $3 \cdot 3 - 5 = 9 - 5 = 4$ Ratios: $2/1 = 2$, $4/2 = 2$ (consistent, valid) - For $x=1.5$: Terms: $1.5 - 2 = -0.5$, $1.5 - 1 = 0.5$, $3 \cdot 1.5 - 5 = 4.5 - 5 = -0.5$ Ratios: $0.5 / (-0.5) = -1$, $(-0.5) / 0.5 = -1$ (consistent, valid) **Final answer:** $$\boxed{x = 3 \text{ or } x = 1.5}$$