Three members GP

The sum of three numbers in GP (geometric progression) is 21, and the sum of their squares is 189. Find the numbers.

Final Answer:

a = 3
b = 6
c = 12
a₂ = 12
b₂ = 6
c₂ = 3

a + b + c = 21 a^{2} + b^{2} + c^{2} = 189 b = q a c = q^{2} a a + q a + q^{2} a = 21 a^{2} + q^{2} a^{2} + q^{4} a^{2} = 189 a (1 + q + q^{2}) = 21 a^{2} (1 + q^{2} + q^{4}) = 189 2 1^{2} (1 + q^{2} + q^{4}) = 189 (1 + q + q^{2})^{2} 252 q^{4} - 378 q^{3} - 126 q^{2} - 378 q + 252 = 0 q^{4} + q^{2} + 1 = (q^{2} - q + 1) \cdot (q^{2} + q + 1) 2 1^{2} (q^{2} - q + 1) \cdot (q^{2} + q + 1) = 189 (1 + q + q^{2})^{2} 2 1^{2} (q^{2} - q + 1) = 189 (1 + q + q^{2}) 2 1^{2} (q^{2} - q + 1) = 189 (1 + q + q^{2}) 252 q^{2} - 630 q + 252 = 0 252 = 2^{2} \cdot 3^{2} \cdot 7 630 = 2 \cdot 3^{2} \cdot 5 \cdot 7 GCD (252, 630, 252) = 2 \cdot 3^{2} \cdot 7 = 126 2 q^{2} - 5 q + 2 = 0 a = 2; b = - 5; c = 2 D = b^{2} - 4 a c = 5^{2} - 4 \cdot 2 \cdot 2 = 9 D > 0 q_{1, 2} = \frac{- b \pm D}{2 a} = \frac{5 \pm 9}{4} q_{1, 2} = \frac{5 \pm 3}{4} q_{1, 2} = 1.25 \pm 0.75 q_{1} = 2 q_{2} = 0.5 a = 21 / (1 + q_{1} + q_{1}^{2}) = 21 / (1 + 2 + 2^{2}) = 3

b = q_{1} \cdot a = 2 \cdot 3 = 6

c = q_{1} \cdot b = 2 \cdot 6 = 12

a_{2} = 21 / (1 + q_{2} + q_{2}^{2}) = 21 / (1 + 0.5 + 0 . 5^{2}) = 12

b_{2} = q_{2} \cdot a_{2} = 0.5 \cdot 12 = 6

c_{2} = q_{2} \cdot b_{2} = 0.5 \cdot 6 = 3

Help us improve! If you spot a mistake, please let let us know. Thank you!

Tips for related online calculators

Are you looking for help with calculating roots of a quadratic equation?
Do you have a linear equation or system of equations and are looking for a solution? Or do you have a quadratic equation?

algebra

arithmetic

Grade of the word problem