A: \( 4 x+2 ; 2 x-6 ; x-8 \) \( \begin{array}{ll}\text { C: } 7 x+1 ; x+3 ; x-5 & \text { D: } x-1 ; 2 x-14 ; 3 x \\ 2.1 & \text { Determine the value(s) of } x \text { for which these terms form geomet } \\ & \text { sequences. }\end{array} \)

To determine the values of \( x \) for which each given sequence forms a geometric sequence, we'll analyze each sequence individually. A geometric sequence satisfies the condition that the ratio of consecutive terms is constant. For three terms \( a \), \( b \), and \( c \) to form a geometric sequence, the following condition must hold: \[ \frac{b}{a} = \frac{c}{b} \] Let's apply this condition to each sequence: --- ### **Sequence A: \( 4x + 2 \), \( 2x - 6 \), \( x - 8 \)** **Condition:** \[ \frac{2x - 6}{4x + 2} = \frac{x - 8}{2x - 6} \] **Cross-Multiplying:** \[ (2x - 6)^2 = (4x + 2)(x - 8) \] **Expanding Both Sides:** \[ 4x^2 - 24x + 36 = 4x^2 - 30x - 16 \] **Simplifying:** \[ -24x + 36 = -30x - 16 \\ 6x = -52 \\ x = -\frac{26}{3} \] **Verification:** - **Term 1:** \( 4\left(-\frac{26}{3}\right) + 2 = -\frac{104}{3} + \frac{6}{3} = -\frac{98}{3} \neq 0 \) - **Term 2:** \( 2\left(-\frac{26}{3}\right) - 6 = -\frac{52}{3} - \frac{18}{3} = -\frac{70}{3} \neq 0 \) **Conclusion:** **\( x = -\frac{26}{3} \)** makes Sequence A a geometric sequence. --- ### **Sequence C: \( 7x + 1 \), \( x + 3 \), \( x - 5 \)** **Condition:** \[ \frac{x + 3}{7x + 1} = \frac{x - 5}{x + 3} \] **Cross-Multiplying:** \[ (x + 3)^2 = (7x + 1)(x - 5) \] **Expanding Both Sides:** \[ x^2 + 6x + 9 = 7x^2 - 34x - 5 \] **Simplifying:** \[ 0 = 6x^2 - 40x - 14 \\ 3x^2 - 20x - 7 = 0 \] **Solving the Quadratic Equation:** \[ x = \frac{20 \pm \sqrt{400 + 84}}{6} = \frac{20 \pm 22}{6} \] **Solutions:** \[ x = \frac{42}{6} = 7 \\ x = \frac{-2}{6} = -\frac{1}{3} \] **Verification:** - **For \( x = 7 \):** - **Term 1:** \( 7(7) + 1 = 50 \neq 0 \) - **Term 2:** \( 7 + 3 = 10 \neq 0 \) - **For \( x = -\frac{1}{3} \):** - **Term 1:** \( 7\left(-\frac{1}{3}\right) + 1 = -\frac{4}{3} \neq 0 \) - **Term 2:** \( -\frac{1}{3} + 3 = \frac{8}{3} \neq 0 \) **Conclusion:** **\( x = 7 \)** and **\( x = -\frac{1}{3} \)** make Sequence C a geometric sequence. --- ### **Sequence D: \( x - 1 \), \( 2x - 14 \), \( 3x \)** **Condition:** \[ \frac{2x - 14}{x - 1} = \frac{3x}{2x - 14} \] **Cross-Multiplying:** \[ (2x - 14)^2 = (x - 1)(3x) \] **Expanding Both Sides:** \[ 4x^2 - 56x + 196 = 3x^2 - 3x \] **Simplifying:** \[ x^2 - 53x + 196 = 0 \] **Solving the Quadratic Equation:** \[ x = \frac{53 \pm \sqrt{2809 - 784}}{2} = \frac{53 \pm 45}{2} \] **Solutions:** \[ x = \frac{98}{2} = 49 \\ x = \frac{8}{2} = 4 \] **Verification:** - **For \( x = 49 \):** - **Term 1:** \( 49 - 1 = 48 \neq 0 \) - **Term 2:** \( 2(49) - 14 = 84 \neq 0 \) - **For \( x = 4 \):** - **Term 1:** \( 4 - 1 = 3 \neq 0 \) - **Term 2:** \( 2(4) - 14 = -6 \neq 0 \) **Conclusion:** **\( x = 49 \)** and **\( x = 4 \)** make Sequence D a geometric sequence. --- ### **Summary of Solutions** - **Sequence A:** \( x = -\dfrac{26}{3} \) - **Sequence C:** \( x = 7 \) and \( x = -\dfrac{1}{3} \) - **Sequence D:** \( x = 49 \) and \( x = 4 \)