Question

What are the explicit equation and domain for a geometric sequence with a first term of 4 and a second term of -8? (5 points)

an = 4(-2)n - 1; all integers where n ≥ 0

an = 4(-2)n - 1; all integers where n ≥ 1

an = 4(-12)n - 1; all integers where n ≥ 1

an = 4(-12)n - 1; all integers where n ≥ 0

Answer 1

i know it's a or b but i don't know how to find the domain

Answer 2

what do you think the common ratio is...?

so

4 x r = -8

r = ?

Answer 3

i told you.. I know it's a or b i just don't know the domain

Answer 4

wait no isn't the common ratio -8/4

Answer 5

bc it's 2nd term divided by first

Answer 6

ok... so if you know its a or b

look at a and substitute n = 0 into the equation.... what do you get..?

Answer 7

well the common ratio is the value you multiply the 1st term by to get the 2nd etc...

it can be found by dividing the 2nd term by the 1st term...

Answer 8

but what happens when you substitute n = 0

Answer 9

how do i find the domain is what I'm asking

Answer 10

nvm

Answer 11

8i got your message late lol

Answer 12

so is it b

Answer 13

well to find the domain, you have 2 choices... so substitute each choice

n =0 into the equation and then n = 1 and see which gives the 1st term of 4

Answer 14

so does

\[a_{0} = 4\times (-2)^{0 -1}\]

is

\[A_{0} = 4\]

Answer 15

oh so it's a

Answer 16

thank you

Answer 17

umm it was b

Answer 18

lol... no its not a if you do the calculation its

\[a_{0} = 4 \times (-2)^{-1} = 4 \times \frac{-1}{2} = -2\]

all you needed to do was read the question, you were told

\[a_{1} = 4 \]

the 1st term was 4 so n = 1

\[a_{1} = 4 \times (-2)^{1 -1} = 4 \times (-2)^0 = 4 \times 1 =4\]