## anonymous 4 years ago Identify the 12th term of a geometric sequence where a1 = 8 and a6 = –8,192.

1. anonymous

a6 = a1*r^5 -8192 = 8*r^5 r^5 = -1024 r = -2 so a12 = a1*r^11 = 8*(-2)^11 = -16384 i guess

2. anonymous

3. anonymous

ahh I see now I got dizzy again lol it's too late here r = -4

4. anonymous

so a12 = 8*(-4)^11 sorry for many mistakes lol

5. anonymous

ahh, thats not good!:(

6. anonymous

but be sorry you have no idea how helpful you are(:

7. anonymous

yess thats right!

8. anonymous

what about Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth.

9. anonymous

a5 = a1*r^4 150.6 = 16*r^4 so u can figure r by using calculator 17th term to the nearest hundredth?? I don't understand my eng is not good enough lol

10. anonymous

haha ummm dont worry about that part? so like Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term

11. campbell_st

stephanizer you need to evaluate the answer for your 1st term $T _{12} = 8\times(-4)^{11}$ as the maths is correct.

12. anonymous

ok r^4 is approximately 6.4 so a17 = a1*r^16 = a1*((r^4))^4 = 8*(6.4)^4=.....

13. anonymous

but I want to understand eng too lol

14. anonymous

no no r is approximately 9.4 lol

15. campbell_st

I got $150.6 = 16r^4$ r = 1.75 then $T _{17} = 16 \times(1.75)^{16} =123802$

16. anonymous

^^ how you figure r out ?? did you use calculator??

17. campbell_st

$r^4 = \frac{150.6}{16}$ then $r = \sqrt[4]{9.4125} = 1.7499$