alexwing285
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The first three terms of a geometric sequence are shown below.

What is the eighth term of the sequence?
A. -128x^8-384x^7
B. 128x^8+384x^7
C. 256x^9+768x^8
D. -256x^9-768x^8

The first three terms of a geometric sequence are shown below What is the eighth term of the sequence A 128x8384x7 B 128x8384x7 C 256x9768x8 D 256x9768x8 class=

Respuesta :

irspow
This is a geometric sequence of the form:

a(n)=(x+3)(-2x)^(n-1)  because each term is -2x times the previous term...so

a(8)=(x+3)(-128x^7)

a(8)=-128x^8-384x^7

Answer:

[tex]a_8=-128x^8-384x^{7}[/tex]

Step-by-step explanation:

G.P. = [tex]x+3,-2x^2-6x,4x^3+12x^2+...[/tex]

So, first term = a=x+3

Common ratio = [tex]r =\frac{a_2}{a_1}[/tex]

                       = [tex]\frac{-2x^2-6x}{x+3}[/tex]

                       = [tex]\frac{-2x(x+3)}{x+3}[/tex]

                       = [tex]-2x[/tex]

So, r = -2x

nth term of G.P. = [tex]a_n=ar^{n-1}[/tex]

Substitute n = 8

[tex]a_8=(x+3)(-2x)^{8-1}[/tex]

[tex]a_8=(x+3)(-2x)^{7}[/tex]

[tex]a_8=(x+3)(-128)(x)^{7}[/tex]

[tex]a_8=-128x^8-384x^{7}[/tex]

Hence the eighth term of the sequence is  [tex]a_8=-128x^8-384x^{7}[/tex]

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