Question

PLEASE HELP these are the last to problems on my homework. PLEASE PLEASE HELP!!!!!!!!
What is the 8th term of the geometric sequence where a_1 = 1024 and a_3 = 64?


What is the sum of a 6–term geometric series if the first term is 8 and the last term is 134,456?

Answer 1

in your first question if a_1 = 1024 and a_3 = 64 then your a_2 = 1024r^2 = 64
square root it...
32r = 8
r = 1/4

so a_8 = a_1r^n-1

a_8 = 1024(1/4)^7

too lazy to do it...but try inputting it in the calculator :DDD

Answer 2

64=1024r^2
64/1024=r^2
1/4=r
\[a _{8}=1024(\frac{1}{4})^7\]

Answer 3

\[\S _{n}=\frac{a _{1}(1-r^n)}{1-r}=\frac{8(1-7^6)}{1-7}\]

Answer 4

do any of you know the second question

Answer 5

Do you even read what people post?

Answer 6

You ask for help. Someone does the problem and posts it and then you ask if anyone knows how to do the problem.

Answer 7

@Mertsj...why did you do it 8(1-7^6)? your r is 1/4 right?