Question

Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -12 and 768, respectively.

Answer 1

@satellite73 Help please!

Answer 2

your in good hands here

Answer 3

Find the value of the variable. If your answer is not an integer, leave it in simplest radical form.

Answer 4

That just makes me think of the statefarm commercial.

Answer 5

that's ALL STATE

Answer 6

Right, right....my bad D:

Answer 7

like a good neighbour, @SithsAndGiggles is there

Answer 8

:D

Answer 9

haha right

Answer 10

jeez i wonder what he is writing ...

Answer 11

can either one of yall help me with this?

Answer 12

i think it's been like 5+ minutes lol i hope it's good :D

Answer 13

@ThatHaley_Girl post in another thread and i will answer it

Answer 14

You should post it as it's own question @ThatHaley_Girl

Answer 15

He could've fallen asleep at the keyboard and his face is just on the keys.

Answer 16

lol yeah

Answer 17

\[a, ar, ar^2, ar^3, ar^4\]
\[a, -12, ar^2, ar^3, 768\] so
\[\frac{ar^4}{ar}=r^3=\frac{768}{-12}=-64\] now find \(r\) then \(a\)

Answer 18

This might be a roundabout method, but here goes.

A geometric sequence has the form \(\left\{a_n\right\}_{n=1}^k=\left\{br^{n-1}\right\}_{n=1}^k\).

\(a_2=-12~\Rightarrow~br=-12\\
a_5=768~\Rightarrow~br^4=768\)

Solving the first equation for \(b\):
\[b=-\frac{12}{r}\]
Substituting into the second,
\[\left(-\frac{12}{r}\right)r^4=768\\
-12r^3=768\\
r^3=64\\
r=4~~~\text{(excluding complex solutions)}\]
Now you can solve for \(b\) by substituting this into either equation, then you get the general sequence \(a_n\).

Answer 19

Sorry for the wait guys ;)

Answer 20

yeah, but after all that \(r\) is negative right?

Answer 21

\[r^3=-64\iff r=-4\]

Answer 22

Ah, yes, I left out the minus sign! Thanks for pointing it out @satellite73

Answer 23

hmmm my answer choices are these:
an = 3 • (-4)n + 1

an = 3 • 4n - 1

an = 3 • (-4)n - 1

an = 3 • 4n
so doesn't the first term have to be 3?
@SithsAndGiggles

Answer 24

we have \(r=-4\) and \(ar=-12\) telling you \(a=3\)

Answer 25

Yes it is ^^^

Answer 26

therefore it is
\[3\times (-4)^{n-1}\]

Answer 27

oh ok got it was confused for a sec there! Thanks a bunch guys

Answer 28

as usual, it is C
it is always C

Answer 29

we only found \(r\) in all our rambling

Answer 30

lol yep it's always C.

Answer 31

Just because of that, we need a mcq where its a. b. d. e. and no c option.