# 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:

${b}_{1}=80\mathrm{/}\left(1+{q}_{2}+{q}_{2}^{2}+{q}_{2}^{3}\right)=80\mathrm{/}\left(1+\left(-3\right)+\left(-3{\right)}^{2}+\left(-3{\right)}^{3}\right)=-4$

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