Question

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

Answer 1

for a geometric sequence, you have to find the common ration

Answer 2

Would the common ratio be -0.015625?

Answer 3

-4096 = 64r^3
-4096 = 262144r
r = 64?

Answer 4

That's not right...

Answer 5

Help, anyone?

Answer 6

i think
nth term is written as
\[a _{n} = a*r ^{n-1}\]
where r = common ratio a = first term
put n=4
u have been given the value of fourth term as well as the first term

Answer 7

I think the 4th term is 64, it says a4 = 64...not sure if that's the same thing

Answer 8

yeah fourth term is 64

Answer 9

what is the common ratio?

Answer 10

first term can't be negative because we cannot determine the cube root of negative number. I feel something is wrong with the ques

Answer 11

Idk...straight out of the book

Answer 12

sorry cube rot can be negative

Answer 13

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

Answer 14

solve this expression u will get common ratio

Answer 15

\[ a_n = a_1r^{n-1}\]
where n starts at 1
If we solve for a0 and r we can find any term. Since we know two points, its possible to solve.

From the problem, we can see
\[ a_1 = -4095\]

Then
\[a_4 = a_1*r^{4-1}= -4095(r)^{3} = 64 \]

Solving for r:
\[ r = \left (\frac{64}{-4095} \right )^{1/3} \]

Cubed roots can be negative. So r is a negative ratio, it will be an alternating sereies.

\[ r = -\frac{4}{16} = -\frac{1}{4} \]

Now the sequence is defined, plug in n = 6 and solve.

Answer 16

64 = -4096r^3
64 = -68719476736r
r = -1073741824
That doesn't seem right...

Answer 17

You must first divide by -4096 and then take the cube root (inverse of 3rd power) to solve r.

Answer 18

So -1/4 is the common ratio?

Answer 19

yup -1/4 is the common ratio

Answer 20

now determine sixth term using
formula for nth term

Answer 21

6=-4096 * -1/4^6−1

6=-4096 * -1/4^5

6=-4096 * -1/4^5

I'm stuck...

Answer 22

\[\frac{ -4096 }{ 4*4*4*4*4 }\]

Answer 23

\[\frac{ -4096 }{ 1024 }\]
\[-4\]

Answer 24

hence sixth term is -4

Answer 25

Thanks, still not sure how you got it...but thanks.

Answer 26

Mind helping me with a different problem?

Answer 27

Given the geometric sequence where a1 = 4 and the common ratio is 3, what is the domain for n?

Answer 28

i just solve the equation

Answer 29

a1 = 4 and r=3
it signifies that it is an increasing gp
so it can goes upto infinity

Answer 30

By definition, a sequence is defined as a collection of elements who can be ordered in accordance with the cardnality of the natural numbers. While n can be any integer, we tend to stick to the natural numbers.

the domain for this sequence, for most sequences you will deal with in that class, is the natural numbers

\[ n \in \mathbb{N} = \left \{1,2,3,4,5,6,7,8,9,...\right \}\]

Answer 31

Would it be all integers where n ≥ 1?

Or maybe all real numbers?

Answer 32

The input (domain) of a sequence cant be the real numbers, you must be careful. Saying \[n \geq 1 \]

Is too vague, or just plain wrong. It must be a natural number (counting numbers)

Answer 33

The answer choices are:
n ≥ 1
n > 1
n ≥ 4
all real numbers

That's why I thought of n ≥ 1

Answer 34

Out of those choices, the one you chose was the best, but the answers are poorly written.

Answer 35

I got the previous question wrong...

Answer 36

The answer was 4, not -4

Answer 37

Thanks so much- like @bbbbbbbbbbbbbbb said, the correct answer was 4 :)