Question

What is the 6th term of the geometric sequence where a1 = 1,024 and a4 = -16?

Answer 1

@tkhunny

Answer 2

@CGGURUMANJUNATH

Answer 3

First Term = a = 1024
Second Term = ar = ??
Third Term = ar^2 = ??
Fourth Term = ar^3 = -16

Can you solve for r?

Answer 4

umm im not sure how to

Answer 5

It's time to get sure.

You have:

a = 1024
ar^3 = -16

Solve for r.

It may help to observe that 1024 = 2^10 and 16 = 2^4

Answer 6

1073741840

Answer 7

Please demonstrate how you managed that.

a = 1024
ar^3 = -16

Thus, but substitution

1024r^3 = -16

Thus, by division

r^3 = -16/1024 = -(2^4)/(2^10) = -1/(2^6)

Are we getting anywhere?

Answer 8

1024^3=16
then i simplftyed
we are getting somewhere

Answer 9

I see. You should have substitute ONLY for 'a'. The 'r' should not have gone anywhere.

ar^3 = (a)*(r^3)

Substituting for 'a' gives

ar^3 = (a)*(r^3) = 1024(r^3)

Answer 10

oh oj=k

Answer 11

ok

Answer 12

i dont know how to solve it

Answer 13

r^3 = -1/(2^6)

You need to know how to solve that. Give it a go.

Answer 14

r^3=-64

Answer 15

How did 2^6 magically get into the numerator?

What you have shown, so far, is not encouraging. Have you been away from algebra for a while? You may need to consider backing up and doing a little more time in an earlier course.

\(r^3 = -1/(2^6)\)

\(\left(r^3\right)^{1/3} = \left(-1/(2^6)\right)^{1/3}\)

\(r^{3\cdot(1/3)} = (-1)^{1/3}/(2^6)^{1/3}\)

\(r = -1/\left(2^{6\cdot (1/3)}\right)\)

\(r = -1/\left(2^{2}\right) = -1/4\)

There should not be anything in there that you cannot do. If there is, please give that earlier course some thought.