Determine 3948

Determine the quotient and the first member of GP if a3 = 0.52 and a1 + a2 = 0.39.

Correct answer:

q₁ = 2
q₂ = -0.6667
a₁ = 0.13
b₁ = 1.17

Step-by-step explanation:

a_{1} + a_{1} q = 0.39 a_{1} (1 + q) = 0.39 a_{1} = 0.39 / (1 + q) a_{1} \cdot q^{2} = 0.52 0.39 / (1 + q) \cdot q^{2} = 0.52 q^{2} = (1 + q) \cdot 0.52 / 0.39 q^{2} = (1 + q) \cdot 0.52 / 0.39 q^{2} = (1 + q) \cdot 0.52 / 0.39 q^{2} - 1.333 q - 1.333 = 0 a = 1; b = - 1.333; c = - 1.333 D = b^{2} - 4 a c = 1.33 3^{2} - 4 \cdot 1 \cdot (- 1.333) = 7.1111111111 D > 0 q_{1, 2} = \frac{- b \pm D}{2 a} = \frac{1 . 3 3 \pm 7 . 1 1}{2} q_{1, 2} = 0.666667 \pm 1.333333 q_{1} = 2 q_{2} = - 0.666666667 q_{1} = 2

Our quadratic equation calculator calculates it.

q_{2} = (- 0.6667) = - 0.6667

a_{1} = 0.39 / (1 + q_{1}) = 0.39 / (1 + 2) = \frac{1 3}{1 0 0} = 0.13 a_{2} = q_{1} \cdot a_{1} = 2 \cdot 0.13 = \frac{1 3}{5 0} = 0.26 a_{3} = q_{1} \cdot a_{2} = 2 \cdot 0.26 = \frac{1 3}{2 5} = 0.52 s = a_{1} + a_{2} = 0.13 + 0.26 = \frac{3 9}{1 0 0} = 0.39 a_{1} = 0.13

b_{1} = 0.39 / (1 + q_{2}) = 0.39 / (1 + (- 0.6667)) = \frac{1 1 7}{1 0 0} = 1 \frac{1 7}{1 0 0} = 1.17 b_{2} = q_{2} \cdot b_{1} = (- 0.6667) \cdot 1.17 = - \frac{3 9}{5 0} = - 0.78 b_{3} = q_{2} \cdot b_{2} = (- 0.6667) \cdot (- 0.78) = \frac{1 3}{2 5} = 0.52 s = b_{1} + b_{2} = 1.17 + (- 0.78) = \frac{3 9}{1 0 0} = 0.39 b_{1} = 1.17

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high school

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