Question

What is the 8th term of the geometric sequence where a1 = 256 and a3 = 16?

Answer 1

divide off the 256 to get a1 to a 1 and the rest should fall into suit

Answer 2

in other words:

\[r = \sqrt{\frac{16}{256}}\]
\[a8 = r^7\]

Answer 3

i still dont get it srry

Answer 4

there are afew underlying concepts to understand; youd have to dialogue with me in order for me to see where it is you need the help

Answer 5

use the formula
\[a_{n} = a_{1}*r^{n-1}\]

Answer 6

o yeah, i forgot about the 256 i lopped off :)