Sequences AP + GP

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

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

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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