Respuesta :

A recursive rule for the geometric sequence is:
a1=10
an=-8 (an-1)

Solution:

Sequence:
10, -80, 640, -5120

a1=10
a2=-80
a3=640
a4=-5120

a2/a1=(-80)/(10)→a2/a1=-8
a3/a2=(640)/(-80)→a3/a2=-8
a4/a3=(-5120)/(640)→a4/a3=-8

Then:
(an)/(an-1)=-8

Solving for an:
an=(-8)(an-1)
an=-8 (an-1)
frika
From the task data 
[tex]a_1=10 \\ a_2=-80 \\ a_3=640 \\ a_4=-5120[/tex].

If these four terms form the geometrical sequence, then each next term is obtained from the previous by multiplying by the same number q. 

Let's find q:[tex]a_2=a_1\cdot q \Rightarrow q= \frac{a_2}{a_1}= \frac{-80}{10} =-8 [/tex] (you can easily check that q is also [tex] \frac{a_3}{a_2} [/tex] and [tex] \frac{a_4}{a_3} [/tex] ).

Then then-th term of geometrical sequence may be represented as [tex]a_n=a_1q^{n-1}[/tex]. That's why [tex]a_n=10\cdot (-8)^{n-1}[/tex].

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