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.


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


a1+qa1+q2 a1+a3 a1=80 q3 a1=9 qa1 q2=9 q1=9=3 q2=9=3 80=a1(1+q+q2+q3) a1=80/(1+q1+q12+q13)=80/(1+3+32+33)=2a_{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
a2=q1 a1=3 2=6a_{2}=q_{1} \cdot \ a_{1}=3 \cdot \ 2=6
a3=q1 a2=3 6=18a_{3}=q_{1} \cdot \ a_{2}=3 \cdot \ 6=18
a4=q1 a3=3 18=54a_{4}=q_{1} \cdot \ a_{3}=3 \cdot \ 18=54
b1=80/(1+q2+q22+q23)=80/(1+(3)+(3)2+(3)3)=4b_{1}=80 / (1+q_{2}+q_{2}^2+q_{2}^3)=80 / (1+(-3)+(-3)^2+(-3)^3)=-4
b2=q2 b1=(3) (4)=12b_{2}=q_{2} \cdot \ b_{1}=(-3) \cdot \ (-4)=12
b3=q2 b2=(3) 12=36b_{3}=q_{2} \cdot \ b_{2}=(-3) \cdot \ 12=-36
b4=q2 b3=(3) (36)=108b_{4}=q_{2} \cdot \ b_{3}=(-3) \cdot \ (-36)=108

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