Arithmetic geometric sequence

Determine four numbers so that the first three form the successive three terms of an arithmetic sequence with difference d=-3 and the last three form the next terms of a geometric sequence with quotient q=one half.

Final Answer:

a₁ =  9
a₂ =  6
a₃ =  3
a₄ =  1.5

Step-by-step explanation:

$$\begin{gathered} a_2=a_1+d \\ {a}_3=a_2+d \\ d=-3 \\ q=1/2=\frac{1}{2}=0.5 \\ {a}_3=q\cdot a_2 \\ {a}_4=q\cdot a_3 \\ {a}_2=a_1+d \\ {a}_3=a_2+d \\ {a}_3=q\cdot a_2 \\ {a}_2=a_1+(-3) \\ {a}_3=a_2+(-3) \\ {a}_3=0.5\cdot a_2 \\ {a}_1-a_2=3 \\ {a}_2-a_3=3 \\ 0.5a_2-a_3=0 \\ Row3-0.5\cdot Row2\to Row3 \\ -0.5a_3=-1.5 \\ {a}_3=\frac{-1.5}{-0.5}=3 \\ {a}_2=3+a_3=3+3=6 \\ {a}_1=3+a_2=3+6=9 \\ {a}_1=9 \\ {a}_2=6 \\ {a}_3=3 \end{gathered}$$

$a_4=q\cdot a_3=0.5\cdot 3=1.5$

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