## anonymous one year ago 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

1. anonymous

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

2. campbell_st

what do you think the common ratio is...? so 4 x r = -8 r = ?

3. anonymous

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

4. anonymous

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

5. anonymous

bc it's 2nd term divided by first

6. campbell_st

ok... so if you know its a or b look at a and substitute n = 0 into the equation.... what do you get..?

7. campbell_st

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...

8. campbell_st

but what happens when you substitute n = 0

9. anonymous

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

10. anonymous

nvm

11. anonymous

8i got your message late lol

12. anonymous

so is it b

13. campbell_st

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

14. campbell_st

so does $a_{0} = 4\times (-2)^{0 -1}$ is $A_{0} = 4$

15. anonymous

oh so it's a

16. anonymous

thank you

17. anonymous

umm it was b

18. campbell_st

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

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