Question

@jdoe0001

Answer 1

Find the first six terms of the sequence.

a1 = -6, an = 4 • an-1

Answer 2

well... seems you're given everything pretty much \(\bf a_{\color{red}{ n}}=a_1+({\color{red}{ n}}-1)d
\qquad
a_1=-6\qquad a_{\color{red}{n}}=4\cdot a_{\color{red}{n}-1}\implies d=4\)

Answer 3

so the equation will be \(a_{\color{red}{ n}}=-6+({\color{red}{ n}}-1)4\implies
\begin{array}{ccllll}
term&value
\\\hline\\
1&a_{\color{red}{ 1}}=-6+({\color{red}{ 1}}-1)4\\
2&a_{\color{red}{ 1}}=-6+({\color{red}{ 2}}-1)4\\
3&a_{\color{red}{ 1}}=-6+({\color{red}{ 3}}-1)4\\
4&a_{\color{red}{ 1}}=-6+({\color{red}{ 4}}-1)4\\
5&a_{\color{red}{ 1}}=-6+({\color{red}{ 5}}-1)4\\
6&a_{\color{red}{ 1}}=-6+({\color{red}{ 6}}-1)4\\
\end{array}\)

Answer 4

okay so a_n=-6+(n-1)4 so .-. you did that in that amount of time.... i just finished writing the answer....

Answer 5

hmm shoot... missed something =) \({\bf a_{\color{red}{ n}}=-6+({\color{red}{ n}}-1)4}\implies
\begin{array}{llll}
term&value
\\\hline\\
1&a_{\color{red}{ 1}}=-6+({\color{red}{ 1}}-1)4\\
2&a_{\color{red}{ 2}}=-6+({\color{red}{ 2}}-1)4\\
3&a_{\color{red}{ 3}}=-6+({\color{red}{ 3}}-1)4\\
4&a_{\color{red}{ 4}}=-6+({\color{red}{ 4}}-1)4\\
5&a_{\color{red}{ 5}}=-6+({\color{red}{ 5}}-1)4\\
6&a_{\color{red}{ 6}}=-6+({\color{red}{ 6}}-1)4\\
\end{array}\)

Answer 6

so itd be -6, -2, 2, 6, 10, 14?

Answer 7

yeap

Answer 8

lol well thats not one of my answers XD

Answer 9

0, 4, -24, -20, -16, -12

-24, -96, -384, -1536, -6144, -24,576

-6, -24, -20, -16, -12, -8

-6, -24, -96, -384, -1536, -6144

Answer 10

hmmmm..... ahemm hold... the mayo .... I think I got the wrong additive =)

Answer 11

I kinda used the 4 as multiplier, implying a geometric sequence, rather than arithmetic

Answer 12

whats an additive? .-.

Answer 13

the difference

Answer 14

-.- i hate math additive seems like you would be adding something.... but when you subtract you get the difference.....

Answer 15

wait so if you dont multiply it by 4 are you dividing it?

Answer 16

well, is an arithmetic sequence... so is meant to have a value added for each term
as opposed to multiplied, that would make it a geometric sequence

Answer 17

okay so its a_n= -6+(n-1)+4?

Answer 18

which will be lol this still dosent come out to any of the answers

Answer 19

right

Answer 20

can you post a quick screenshot of the material?

Answer 21

.-. how do you do that?

Answer 22

hmmm wait... ... the question doesn't say is an arithmetic... so I guess it's a geometric sequence

Answer 23

oh my john......

Answer 24

so that means the d = 4... one sec

Answer 25

\( {\bf a_{\color{red}{ n}}=a_1\cdot r^{{\color{red}{ n}}-1}
\qquad
a_1=-6\qquad a_{\color{red}{n}}=4\cdot a_{\color{red}{n}-1}\implies r=4
\\ \quad \\
a_{\color{red}{ n}}=a_1\cdot r^{{\color{red}{ n}}-1}}\implies
\begin{array}{ccllll}
term&value
\\\hline\\
1&a_{\color{red}{ 1}}=a_1\cdot r^{{\color{red}{ 1}}-1}\\
2&a_{\color{red}{ 2}}=a_1\cdot r^{{\color{red}{ 2}}-1}\\
3&a_{\color{red}{ 3}}=a_1\cdot r^{{\color{red}{ 3}}-1}\\
4&a_{\color{red}{ 4}}=a_1\cdot r^{{\color{red}{ 4}}-1}\\
5&a_{\color{red}{ 5}}=a_1\cdot r^{{\color{red}{ 5}}-1}\\
6&a_{\color{red}{ 6}}=a_1\cdot r^{{\color{red}{ 6}}-1}\\
\end{array} \)

Answer 26

oh my god -.- soooooo -6x4^n-1.....
-6,-24, -96 lol okay so its b

Answer 27

well, B doesn't have a -6 though

Answer 28

yeah c: