Geometric seq

Find the third member of geometric progression if a1 + a2 = 36 and a1 + a3 = 90. Calculate its quotient.

Correct result:

c₁ = 81
c₂ = 373248
q₁ = 3
q₂ = 0

Solution:

a + b = 36 b = q a a + c = 90 c = q \cdot b a + q a = 36 a + q^{2} a = 90 a = 36 / (1 + q) 36 + q^{2} * 36 = 90 * (1 + q) 36 + q^{2} \cdot 36 = 90 \cdot (1 + q) 36 q^{2} - 90 q - 54 = 0 a = 36; b = - 90; c = - 54 D = b^{2} - 4 a c = 9 0^{2} - 4 \cdot 36 \cdot (- 54) = 15876 D > 0 q_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a} = \frac{90 \pm \sqrt{15876}}{72} q_{1, 2} = \frac{90 \pm 126}{72} q_{1, 2} = 1.25 \pm 1.75 q_{1} = 3 q_{2} = - 0.5 Factored form of the equation: 36 (q - 3) (q + 0.5) = 0 a_{1} = 36 / (1 + q_{1}) = 36 / (1 + 3) = 9 a_{2} = 36 / (1 + q_{2}) = 36 / (1 + (- 0.5)) = 72 c_{1} = a_{1} \cdot q_{1}^{2} = 9 \cdot 3^{2} = 81

Our quadratic equation calculator calculates it.

c_{2} = a_{2} \cdot a_{2}^{2} = 72 \cdot 7 2^{2} = 373248

q_{1} = 3

q_{2} = 0

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