Question

What is the 9th term of the geometric sequence where a1 = 1024 and a3 = 64?

Answer 1

\[a_1,a _2=a_1r, a_3=a_1r^2, ...\] so you have
\[\frac{a_3}{a_1}=r^2\] and therefore
\[\frac{1024}{64}=r^2\]
\[16=r^2\]solve for \(r\)

Answer 2

1024*k^2=64
k^2=64/1024=1/16
k=1/4
therefore 9th term is 1024/(4^8)=1024/65536=1/64

Answer 3

lol i had it upside down!

\[\frac{64}{1024}=r^2\]
\[\frac{1}{16}=r^2\]
\[r=\frac{1}{4}\]

Answer 4

it either 2,028
1,728
1,452
2,268

Answer 5

actually it is none of these

Answer 6

\[a, ar, ar^2, ar^3, ar^4, ...\]
\[1024,256,64,16,...\] it is a decreasing sequence