\( x-6 ; 2 x-6 ; 5 x+3 \) forms a GP Determine the first three terms

We are given that the three numbers   x – 6, 2x – 6, 5x + 3 form a geometric progression (GP). In a GP the square of the middle term equals the product of the first and third terms. Therefore, we set up the equation   (2x – 6)² = (x – 6)(5x + 3). Step 1. Expand both sides:  Left side:   (2x – 6)² = 4x² – 24x + 36.  Right side:   (x – 6)(5x + 3) = 5x² + 3x – 30x – 18 = 5x² – 27x – 18. Step 2. Set the equation:   4x² – 24x + 36 = 5x² – 27x – 18. Step 3. Rearrange to bring all terms to one side:   4x² – 24x + 36 – 5x² + 27x + 18 = 0   –x² + 3x + 54 = 0. Multiply through by –1 to make the quadratic standard:   x² – 3x – 54 = 0. Step 4. Solve the quadratic equation: The discriminant is   D = (–3)² – 4(1)(–54) = 9 + 216 = 225. Taking the square root gives √225 = 15. Thus, the solutions are:   x = [3 ± 15] / 2. So, the two possible values for x are:   x = (3 + 15)/2 = 18/2 = 9,   x = (3 – 15)/2 = (–12)/2 = –6. Step 5. Find the three terms for each value of x: Case 1: When x = 9  First term: x – 6 = 9 – 6 = 3,  Second term: 2x – 6 = 18 – 6 = 12,  Third term: 5x + 3 = 45 + 3 = 48. Check: 12/3 = 4 and 48/12 = 4, so the common ratio is 4. Case 2: When x = –6  First term: x – 6 = –6 – 6 = –12,  Second term: 2x – 6 = –12 – 6 = –18,  Third term: 5x + 3 = –30 + 3 = –27. Check: (–18)/(–12) = 1.5 and (–27)/(–18) = 1.5, so here the common ratio is 1.5. Thus, the first three terms of the GP can be either:  • 3, 12, 48  (or with common ratio 4), or  • –12, –18, –27  (or with common ratio 1.5). These are the possible sequences that satisfy the given condition.