Question

What is the 6th term of the geometric sequence where a1 = -4,096 and a4 = 64?
1
-1
4
-4

Answer 1

first we need to find the ratio of the geometric series,
we'll do this by writing the equation for the a4 term of a geometric series.

the geometric series equation is
an = a1*r^(n-1)

what are the values for
a1, n, and an if we just look at the given info for a4?

Answer 2

a1 = -4096

Answer 3

good,
keep going though, if we just look at what the terms we know for when n=4

an = a1*r^(n-1)
64 = -4096*r(n-1)
what would n be?

Answer 4

so the 6th term is 4?

Answer 5

mmm....
lets not jump ahead.

look at this:
an = a1*r^(n-1)
if I only look at the a4 term then I know:
an = 64
a1 = -4096
n = 4

take these values and plug them into
an = a1*r^(n-1)

what do you get?

Answer 6

an = a1*r^(n-1)

becomes
64 = -4096 * r^(4-1)
64 = -4096 * r^(3)

the next step would be to solve for r, divide both sides by -4096

Answer 7

I got -0.015 = r^(3) is that right?

Answer 8

yep!

now we need to get rid of the power of 3 on the r, we do this by taking the cube root of both sides,

\[\sqrt[3]{-0.015} = r\]

Answer 9

do u remember how to do cube roots on your calculator?

Answer 10

I got -0.246 = r

Answer 11

yep, just round it up to -0.25
so that is what r equals

plug that into the an=a1*r(n-1)
and solve for a6

Answer 12

I got 4... thank you!!!!:D

Answer 13

I get -1

hmm...

Answer 14

a6 = -4096*(-0.25)^6

Answer 15

So 64 is a4 and I multiplied 64*-0.25 = -16
-16*-0.25 = 4 and 4 would be the 6th term

Answer 16

oh, i see what you did
use this equations:
an = a1 * r^(n-1)
we found r, and know n and a1

Answer 17

You got me confused, so then the answer is -1?

Answer 18

geometric series equation is:
an = a1 * r^(n-1)

we solved for what r is -0.25
n would be 6 since we are looking for the 6th term in the sequence
and a1 was given to use as -4096

and we are solving for an, which in our case would be called a6

remember that the a6 means a 'sub' 6 not a*6