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.

Result

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

Solution:

a+b+c=21 a2+b2+c2=189  b=qa c=q2a  a+qa+q2a=21 a2+q2a2+q4a2=189  a(1+q+q2)=21 a2(1+q2+q4)=189  212(1+q2+q4)=189(1+q+q2)2 252 q4378 q3126 q2378 q+252=0  q4+q2+1=(q2q+1) (q2+q+1)  212(q2q+1) (q2+q+1)=189(1+q+q2)2  212(q2q+1)=189(1+q+q2) 252q2630q+252=0  a=252;b=630;c=252 D=b24ac=63024252252=142884 D>0  q1,2=b±D2a=630±142884504 q1,2=630±378504 q1,2=1.25±0.75 q1=2 q2=0.5   Factored form of the equation:  252(q2)(q0.5)=0   a=21/(1+q1+q12)=21/(1+2+22)=3a+b+c=21 \ \\ a^2+b^2+c^2=189 \ \\ \ \\ b=qa \ \\ c=q^2a \ \\ \ \\ a+qa+q^2a=21 \ \\ a^2 + q^2a^2+q^4a^2=189 \ \\ \ \\ a(1+q+q^2)=21 \ \\ a^2(1 + q^2+q^4)=189 \ \\ \ \\ 21^2(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) \ \\ \ \\ 21^2(q^2 - q + 1) \cdot \ (q^2 + q + 1)=189 (1+q+q^2)^2 \ \\ \ \\ 21^2(q^2 - q + 1)=189 (1+q+q^2) \ \\ 252q^2 -630q +252=0 \ \\ \ \\ a=252; b=-630; c=252 \ \\ D=b^2 - 4ac=630^2 - 4\cdot 252 \cdot 252=142884 \ \\ D>0 \ \\ \ \\ q_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 630 \pm \sqrt{ 142884 } }{ 504 } \ \\ q_{1,2}=\dfrac{ 630 \pm 378 }{ 504 } \ \\ q_{1,2}=1.25 \pm 0.75 \ \\ q_{1}=2 \ \\ q_{2}=0.5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 252 (q -2) (q -0.5)=0 \ \\ \ \\ \ \\ a=21/(1+q_{1}+q_{1}^2)=21/(1+2+2^2)=3

Checkout calculation with our calculator of quadratic equations.

b=q1 a=2 3=6b=q_{1} \cdot \ a=2 \cdot \ 3=6
c=q1 b=2 6=12c=q_{1} \cdot \ b=2 \cdot \ 6=12
a2=21/(1+q2+q22)=21/(1+0.5+0.52)=12a_{2}=21/(1+q_{2}+q_{2}^2)=21/(1+0.5+0.5^2)=12
b2=q2 a2=0.5 12=6b_{2}=q_{2} \cdot \ a_{2}=0.5 \cdot \ 12=6
c2=q2 b2=0.5 6=3c_{2}=q_{2} \cdot \ b_{2}=0.5 \cdot \ 6=3



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