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:

$$\begin{gathered} a_1+q{a}_1+q^2{a}_1+a^3{a}_1=80 \\ {q}^3{a}_1=9q{a}_1 \\ {q}^2=9 \\ {q}_1=\sqrt{9}=3 \\ {q}_2=-\sqrt{9}=-3 \\ 80=a_1(1+q+q^2+q^3) \\ {a}_1=80\mathrm{/}(1+q_1+q_1^2+q_1^3)=80\mathrm{/}(1+3+3^2+3^3)=2 \end{gathered}$$

$a_2=q_1\cdot a_1=3\cdot 2=6$

$a_3=q_1\cdot a_2=3\cdot 6=18$

$a_4=q_1\cdot a_3=3\cdot 18=54$

$b_1=80\mathrm{/}(1+q_2+q_2^2+q_2^3)=80\mathrm{/}(1+(-3)+(-3{)}^2+(-3{)}^3)=-4$

$b_2=q_2\cdot b_1=(-3)\cdot (-4)=12$

$b_3=q_2\cdot b_2=(-3)\cdot 12=-36$

$b_4=q_2\cdot b_3=(-3)\cdot (-36)=108$

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