Question

What is the 6th term of the geometric sequence where a1 = 128 and a3 = 8?

0.03125

0.0625

0.125

0.15625

Answer 1

anybody know what the equation is

Answer 2

a1 is the first term ryt

Answer 3

yes

Answer 4

well a1 = 128, yes

Answer 5

do u know the total number of terms

Answer 6

6 i guess because that is what we are trying to find

Answer 7

\[ar ^{n-1}\]

Answer 8

ok.so our n is 6? or is it just n because thats what we are trying to find.

Answer 9

we are trying to find out r the common difference

Answer 10

would we plug in the info from what we already know to a and n?

Answer 11

just a min

Answer 12

\[a ^{3}=a1xr ^{2}\]

Answer 13

substitute the values of a3 and a1 to find r

Answer 14

@SolomonZelman

Answer 15

i did as this guy said and got 1/4 converted to decimal is.25 but that is not an aswer choice

Answer 16

r=1/4
128,32,8,2,0.5,0.125

Answer 17

the ans is 0.125