Sequences AP + GP

The three numbers that make up the arithmetic sequence have the sum of 30. If we subtract from the first 5, from the second 4 and keep the third, we get the geometric sequence. Find AP and GP members.

Correct answer:

a₁ =  8
b₁ =  10
c₁ =  12
a₂ =  17
b₂ =  10
c₂ =  3
g₁ =  3
g₂ =  6
g₃ =  12
k₁ =  12
k₂ =  6
k₃ =  3

Step-by-step explanation:

$$\begin{gathered} a+b+c=30 \\ b=a+d \\ c=b+d \\ q\cdot (a-5)=(b-4) \\ q\cdot (b-4)=c \\ 3a+3d=30 \\ a=(30-3d)/3=10-d \\ b=a+d=10-d+d \\ b=10 \\ c=10+d \\ q\cdot (10-d-5)=6 \\ q\cdot 6=10+d \\ (5-d)/6=6/(10+d)=\frac{3}{5}=0.6 \\ (5-d)\ast (10+d)=6\ast 6 \\ (5-d)\cdot (10+d)=6\cdot 6 \\ -d^2-5d+14=0 \\ {d}^2+5d-14=0 \\ a=1;b=5;c=-14 \\ D=b^2-4ac=5^2-4\cdot 1\cdot (-14)=81 \\ D>0 \\ {d}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-5\pm \sqrt{81}}{2} \\ {d}_{1,2}=\frac{-5\pm 9}{2} \\ {d}_{1,2}=-2.5\pm 4.5 \\ {d}_1=2 \\ {d}_2=-7 \\ \text{ Factored form of the equation: } \\ (d-2)(d+7)=0 \\ {a}_1=b-d_1=10-2=8 \end{gathered}$$

Our quadratic equation calculator calculates it.

$b_1=b=10$

$c_1=b+d_1=10+2=12$

$a_2=b-d_2=10-(-7)=17$

$b_2=b=10$

$c_2=b+d_2=10+(-7)=3$

$g_1=a_1-5=8-5=3$

$g_2=b-4=10-4=6$

$g_3=c_1=12$

$k_1=a_2-5=17-5=12$

$k_2=b-4=10-4=6$

$$\begin{gathered} k_3=c_2=3 \\ \text{ Verifying Solution: } \\ {Q}_1=g_2/g_1=6/3=2 \\ {Q}_{11}=g_3/g_2=12/6=2 \\ {Q}_2=k_2/k_1=6/12=\frac{1}{2}=0.5 \\ {Q}_{22}=k_3/k_2=3/6=\frac{1}{2}=0.5 \end{gathered}$$

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