Respuesta :
so hmm is a geometric sequence, meaning, the next term is found by multiplying it by "something", namely the "common ratio"
now, if the next term is the product of the common ratio and the previous term, that means, if we divide the previous term by the next term, the quotient will then be the "common ratio", let's do that then
let's divide the 2nd term by the 1st term then
[tex]\bf \cfrac{-2x^2-6x}{x+3}\implies \cfrac{-2x\underline{(x+3)}}{\underline{(x+3)}}\implies \boxed{-2x}\impliedby \textit{common ratio}\\\\ -----------------------------\\\\[/tex]
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{value of first term}\\ r=\textit{common ratio}\\ ----------\\ a_1=x+3\\ n=8\\ r=-2x \end{cases} \\\\\\ a_8=(x+3)(-2x)^{8-1}\implies a_8=(x+3)(-2x)^7 \\\\\\ a_8=(x+3)(-2^7x^7)\implies a_8=(x+3)(-128x^7) \\\\\\ a_8=-128x^8-384x^7[/tex]
now, if the next term is the product of the common ratio and the previous term, that means, if we divide the previous term by the next term, the quotient will then be the "common ratio", let's do that then
let's divide the 2nd term by the 1st term then
[tex]\bf \cfrac{-2x^2-6x}{x+3}\implies \cfrac{-2x\underline{(x+3)}}{\underline{(x+3)}}\implies \boxed{-2x}\impliedby \textit{common ratio}\\\\ -----------------------------\\\\[/tex]
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{value of first term}\\ r=\textit{common ratio}\\ ----------\\ a_1=x+3\\ n=8\\ r=-2x \end{cases} \\\\\\ a_8=(x+3)(-2x)^{8-1}\implies a_8=(x+3)(-2x)^7 \\\\\\ a_8=(x+3)(-2^7x^7)\implies a_8=(x+3)(-128x^7) \\\\\\ a_8=-128x^8-384x^7[/tex]
