Question

The first three terms of a geometric sequence are shown below. x+3,-2x2-6x,4x3+12x2,.... What is the eighth term of the sequence?

Answer 1

Geometric sequence is times r each time
 a term divideded by the previous term is r

(-2x^2-6x)/(x+3)=-2x

first term is x+3
an=a1(r)^(n-1)
an=(x+3)(-2x)^(r-1)
8th term
an=(x+3)(-2x)^(8-1)
an=(x+3)(-2x)^7
an=(x+3)(-128x^7)
an=-128x^7-384 is the 8th term

Answer 2

You have first three terms of the geometrical sequence:

$a_1=x+3,\\ a_2=-2x^2-6x=-2x(x+3),\\a_3=4x^3+12x^2=4x^2(x+3).$

You should know how the terms of geometrical sequence are formed:

$a_2=a_1\cdot r,\\ a_3=a_2\cdot r=a_1\cdot r^2,...\\a_n=a_1\cdot r^{n-1}$

where r is a ratio of geometrical sequence.

Comparing first three given terms, you can conclude that $r=-2x$ and $a_1=x+3$. Then

$a_8=a_1\cdot r^7=(x+3)\cdot (-2x)^7=-128x^7(x+3).$

Answer: $a_8=-128x^7(x+3).$