Question

What is the 8th term of the geometric sequence where a1 = 1024 and a3 = 64?

Answer 1

Hey - so in a geometric sequence, you're multiplying each term by the same thing to get to the next thing
a1 = 1024
a2 = ?
a3 = 64
a4...a8

Answer 2

so what's 64's relation to 1024? How do you get to 64 in two steps?

Answer 3

I am trying to figure that out...

1024/4=256/4=64...

Answer 4

exactly!

Answer 5

so then keep dividing by 4

Answer 6

a3 = 64
a4 = 64/4 = 16
a5 = 16/4 ... and so on

Answer 7

64/4=16/4=4 a4
4/4=1/4=.25 a5
.25/4=.0625/4=.015625 a6

Answer 8

Wait use mathematics here..

Answer 9

...

Answer 10

watch out on a5 ... and keep them in fraction form. will make it easier.

a3 = 64
a4 = 64/4 = 16
a5 = 16/4 = ...

Answer 11

\[a = 1024\]

\[a_3 = 64\]

\[\large 64 = a \cdot r^{3-1} \implies (1024) \cdot (r^2)\]

Find r from it..

Answer 12

I see.

Answer 13

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

Answer 14

Now you have to find 8th term:

\[\large a_8 = a \cdot (r)^{8-1} \implies a \cdot (r)^7\]

Can you find now ??

Answer 15

.0625?

Answer 16

\[\large a_8 = (1024) \times \frac{1}{4^7} \implies \color{green}{ 0.0625}\]

Yes you are right..

Answer 17

Thank You Both!