Question

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

Answer 1

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 2

Answer:

$a_8=-128x^8-384x^{7}$

Step-by-step explanation:

G.P. = $x+3,-2x^2-6x,4x^3+12x^2+...$

So, first term = a=x+3

Common ratio = $r =\frac{a_2}{a_1}$

                       = $\frac{-2x^2-6x}{x+3}$

                       = $\frac{-2x(x+3)}{x+3}$

                       = $-2x$

So, r = -2x

nth term of G.P. = $a_n=ar^{n-1}$

Substitute n = 8

$a_8=(x+3)(-2x)^{8-1}$

$a_8=(x+3)(-2x)^{7}$

$a_8=(x+3)(-128)(x)^{7}$

$a_8=-128x^8-384x^{7}$

Hence the eighth term of the sequence is  $a_8=-128x^8-384x^{7}$