2 years ago

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

1 answer:

8 0

$a_1=x+3\\\\a_2=-2x^2-6x=-2x(x+3)\\\\a_3=4x^3+12x^2=-2x(-2x^2-6x)=-2x[-2x(x+3)]\\\vdots\\\text{The general term of geometric sequence}\ a_n=a_1r^{n-1}\\\\a_1-first\ term\\r-common\ ratio\\\\a_1=x+3,\ r=-2x\\\\\text{Substitute}\\\\a_n=(x+3)(-2x)^{n-1}\\\\8th\ term\to a_8\\\\\text{Substitute}\ n=8\ \text{to the equation of}\ a_n:\\\\a_8=(x+3)(-2x)^{8-1}=(x+3)(-2x)^7=(x+3)(-128x^7)\\\\\text{use distributive property}\\\\a_8=(x)(-128x^7)+(3)(-128x^7)\\\\\boxed{a_8=-128x^8-384x^7}$

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