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:

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)=2
a2=q1 a1=3 2=6
a3=q1 a2=3 6=18
a4=q1 a3=3 18=54
b1=80/(1+q2+q22+q23)=80/(1+(3)+(3)2+(3)3)=4
b2=q2 b1=(3) (4)=12
b3=q2 b2=(3) 12=36
b4=q2 b3=(3) (36)=108



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