Gp - 80

Sum of the first four members of a geometric progression is 80. Determine they if we know that the fourth member is nine times greater than the second.

Correct result:

a₁ = 2
a₂ = 6
a₃ = 18
a₄ = 54
b₁ = -4
b₂ = 12
b₃ = -36
b₄ = 108

Solution:

a_{1} + q a_{1} + q^{2} a_{1} + a^{3} a_{1} = 80 q^{3} a_{1} = 9 q a_{1} q^{2} = 9 q_{1} = \sqrt{9} = 3 q_{2} = - \sqrt{9} = - 3 80 = a_{1} (1 + q + q^{2} + q^{3}) a_{1} = 80 / (1 + q_{1} + q_{1}^{2} + q_{1}^{3}) = 80 / (1 + 3 + 3^{2} + 3^{3}) = 2

a_{2} = q_{1} \cdot a_{1} = 3 \cdot 2 = 6

a_{3} = q_{1} \cdot a_{2} = 3 \cdot 6 = 18

a_{4} = q_{1} \cdot a_{3} = 3 \cdot 18 = 54

b_{1} = 80 / (1 + q_{2} + q_{2}^{2} + q_{2}^{3}) = 80 / (1 + (- 3) + (- 3)^{2} + (- 3)^{3}) = - 4

b_{2} = q_{2} \cdot b_{1} = (- 3) \cdot (- 4) = 12

b_{3} = q_{2} \cdot b_{2} = (- 3) \cdot 12 = - 36

b_{4} = q_{2} \cdot b_{3} = (- 3) \cdot (- 36) = 108

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