Find k

Find k so that the terms k-3, k+1, and 4k-2 form a geometric sequence. Show your solution.

k₁ = 5
k₂ = 1/3

g_{1} = k - 3 g_{2} = k + 1 g_{3} = 4 k - 2 q = q_{2} / q_{1} = q_{3} / q_{2} q_{2}^{2} = q_{1} \cdot q_{3} (k + 1)^{2} = (k - 3) \cdot (4 \cdot k - 2) (k + 1)^{2} = (k - 3) \cdot (4 \cdot k - 2) - 3 k^{2} + 16 k - 5 = 0 3 k^{2} - 16 k + 5 = 0 a = 3; b = - 16; c = 5 D = b^{2} - 4 a c = 1 6^{2} - 4 \cdot 3 \cdot 5 = 196 D > 0 k_{1, 2} = \frac{- b \pm D}{2 a} = \frac{1 6 \pm 1 9 6}{6} k_{1, 2} = \frac{1 6 \pm 1 4}{6} k_{1, 2} = 2.666667 \pm 2.333333 k_{1} = 5 k_{2} = 0.333333333 Verifying Solution: g_{1} = k_{1} - 3 = 5 - 3 = 2 g_{2} = k_{1} + 1 = 5 + 1 = 6 g_{3} = 4 \cdot k_{1} - 2 = 4 \cdot 5 - 2 = 18 r_{1} = g_{2} / g_{1} = 6 / 2 = 3 r_{2} = g_{3} / g_{2} = 18 / 6 = 3

G_{1} = k_{2} - 3 = 0.3333 - 3 = - \frac{8}{3} = - 2 \frac{2}{3} ≐ - 2.6667 G_{2} = k_{2} + 1 = 0.3333 + 1 = \frac{4}{3} = 1 \frac{1}{3} ≐ 1.3333 G_{3} = 4 \cdot k_{2} - 2 = 4 \cdot 0.3333 - 2 = - \frac{2}{3} ≐ - 0.6667 r_{2} = G_{2} / G_{1} = 1.3333 / (- 2.6667) = - \frac{1}{2} = - 0.5 R_{2} = G_{3} / G_{2} = (- 0.6667) / 1.3333 = - \frac{1}{2} = - 0.5 k_{2} = 0.3333 = \frac{1}{3}

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