11 Determine the value(s) of \( x \) and/or \( y \) in each of the following geometric sequences: a \( 2 ; x ; 32 ; \ldots \) b \( x+1 ; x-1 ; x ; \ldots \) c \( 20 ; x ; y ;-\frac{4}{25} \) d \( 7 x+1 ; 2 x+2 ; x-1 ; \ldots \) e \( 7 ; x ; y ; 56 ; \ldots \)

Let's solve each part step by step, identifying the values of \( x \) and/or \( y \) that make each sequence geometric. In a geometric sequence, the ratio between consecutive terms is constant. ### Problem 11a: **Sequence:** \( 2 ; x ; 32 ; \ldots \) **Solution:** For a geometric sequence, the ratio between consecutive terms is constant: \[ \frac{x}{2} = \frac{32}{x} \] Cross-multiply to solve for \( x \): \[ x^2 = 2 \times 32 = 64 \] \[ x = \pm \sqrt{64} = \pm 8 \] **Answer:** \( x = 8 \) or \( x = -8 \) --- ### Problem 11b: **Sequence:** \( x+1 ; x-1 ; x ; \ldots \) **Solution:** Let the common ratio be \( r \). Then: \[ \frac{x - 1}{x + 1} = \frac{x}{x - 1} \] Cross-multiply: \[ (x - 1)^2 = x(x + 1) \] Expand both sides: \[ x^2 - 2x + 1 = x^2 + x \] Subtract \( x^2 \) from both sides: \[ -2x + 1 = x \] \[ 1 = 3x \] \[ x = \frac{1}{3} \] **Answer:** \( x = \dfrac{1}{3} \) --- ### Problem 11c: **Sequence:** \( 20 ; x ; y ; -\dfrac{4}{25} \) **Solution:** Let the common ratio be \( r \). \[ x = 20r \] \[ y = x \cdot r = 20r^2 \] \[ -\frac{4}{25} = y \cdot r = 20r^3 \] Solve for \( r \): \[ 20r^3 = -\frac{4}{25} \] \[ r^3 = -\frac{4}{25 \times 20} = -\frac{4}{500} = -\frac{1}{125} \] \[ r = -\frac{1}{5} \] Now find \( x \) and \( y \): \[ x = 20 \times \left(-\frac{1}{5}\right) = -4 \] \[ y = 20 \times \left(-\frac{1}{5}\right)^2 = 20 \times \frac{1}{25} = \frac{20}{25} = \frac{4}{5} \] **Answer:** \( x = -4 \) and \( y = \dfrac{4}{5} \) --- ### Problem 11d: **Sequence:** \( 7x + 1 ; 2x + 2 ; x - 1 ; \ldots \) **Solution:** Let the common ratio be \( r \). \[ \frac{2x + 2}{7x + 1} = \frac{x - 1}{2x + 2} \] Cross-multiply: \[ (2x + 2)^2 = (7x + 1)(x - 1) \] Expand both sides: \[ 4x^2 + 8x + 4 = 7x^2 - 7x + x - 1 \] Simplify: \[ 4x^2 + 8x + 4 = 7x^2 - 6x - 1 \] Bring all terms to one side: \[ 0 = 7x^2 - 6x - 1 - 4x^2 - 8x - 4 \] \[ 0 = 3x^2 - 14x - 5 \] Solve the quadratic equation: \[ x = \frac{14 \pm \sqrt{(-14)^2 - 4 \times 3 \times (-5)}}{2 \times 3} = \frac{14 \pm \sqrt{196 + 60}}{6} = \frac{14 \pm \sqrt{256}}{6} = \frac{14 \pm 16}{6} \] \[ x = \frac{30}{6} = 5 \quad \text{or} \quad x = \frac{-2}{6} = -\frac{1}{3} \] We need to verify which solution(s) make sense in the context of a geometric sequence (i.e., the ratios are consistent). **For \( x = 5 \):** \[ 7(5) + 1 = 36; \quad 2(5) + 2 = 12; \quad 5 - 1 = 4 \] Ratios: \[ \frac{12}{36} = \frac{1}{3}; \quad \frac{4}{12} = \frac{1}{3} \] Consistent. **For \( x = -\frac{1}{3} \):** \[ 7\left(-\frac{1}{3}\right) + 1 = -\frac{7}{3} + 1 = -\frac{4}{3}; \quad 2\left(-\frac{1}{3}\right) + 2 = -\frac{2}{3} + 2 = \frac{4}{3}; \quad -\frac{1}{3} - 1 = -\frac{4}{3} \] Ratios: \[ \frac{\frac{4}{3}}{-\frac{4}{3}} = -1; \quad \frac{-\frac{4}{3}}{\frac{4}{3}} = -1 \] Consistent. **Answer:** \( x = 5 \) or \( x = -\dfrac{1}{3} \) --- ### Problem 11e: **Sequence:** \( 7 ; x ; y ; 56 ; \ldots \) **Solution:** Let the common ratio be \( r \). \[ x = 7r \] \[ y = x \cdot r = 7r^2 \] \[ 56 = y \cdot r = 7r^3 \] Solve for \( r \): \[ 7r^3 = 56 \] \[ r^3 = 8 \] \[ r = 2 \] Now find \( x \) and \( y \): \[ x = 7 \times 2 = 14 \] \[ y = 7 \times 2^2 = 7 \times 4 = 28 \] **Answer:** \( x = 14 \) and \( y = 28 \) --- ### Summary of Answers: a) \( x = 8 \) or \( x = -8 \) ---