# Thread: Help with finding the formula to find the nth term of a geometric sequence.

1. ## Help with finding the formula to find the nth term of a geometric sequence.

Geometric sequence is 3/2, -3/4, and 3/8 are the first three terms. what represents the nth term of the sequence?

I know a is the first term, 3/2.
ratio is just a term divided by the term proceeding it.
ar^(n-1) is how you find a specific term.

I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html

says the final answer is:
a(n) = 3*(-1)^(n-1)/2^(n)

My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?

2. Originally Posted by Johnny_L
I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

the final answer the book says...is:
a(n) = 3*(-1)^(n-1)/2^(n)

My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?
What did you get when you multiplied things out? You split the 3 from the /2 to get (3)(1/2), you noted that (-1/2)^(n-1) = (-1)^(n-1) / 2^(n-1) = (-1)^(n-1) * (1/2)^(n-1), and then you noted that you've now got a (1/2) = (1/2)^1 and a (1/2)^(n-1), so....

3. Originally Posted by Johnny_L
I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html

says the final answer is:
a(n) = 3*(-1)^(n-1)/2^(n)
What you got is correct; final answer also correct:
simply another way to get same results.

Standard formula is ar^(n-1) which you got...

4. Originally Posted by stapel
What did you get when you multiplied things out? You split the 3 from the /2 to get (3)(1/2), you noted that (-1/2)^(n-1) = (-1)^(n-1) / 2^(n-1) = (-1)^(n-1) * (1/2)^(n-1), and then you noted that you've now got a (1/2) = (1/2)^1 and a (1/2)^(n-1), so....
(3/2)(-1/2)^(n-1)

3 X-1 = -3; 2 X 2 =4

3/4^(n-1)

why aren't we multiplying numerator by numerator and denominator by denominator?

I don't understand how we get to this

a(n) = 3*(-1)^(n-1)/2^(n)

form at all...

5. Originally Posted by Johnny_L
(3/2)(-1/2)^(n-1)

3 X-1 = -3; 2 X 2 =4

3/4^(n-1)
(Use * for multiplication symbol...now standard)

Nooooooo: (3/2) * [(-1/2)^(n-1)] ; exponent first, then multiplication...

6. Originally Posted by Johnny_L
Geometric sequence is 3/2, -3/4, and 3/8 are the first three terms. what represents the nth term of the sequence?

I know a is the first term, 3/2.
ratio is just a term divided by the term proceeding it.
ar^(n-1) is how you find a specific term.

I get all the way up to here: a(n) = ar^(n-1) = (3/2)(-1/2)^(n-1)

the final answer the book says, and this website I got help from: http://www.algebra.com/algebra/homew...on.391759.html

says the final answer is:
a(n) = 3*(-1)^(n-1)/2^(n)

My question is, what happens to the other 2 in the denominator, and where does that n exponent in the denominator come from?
This is not hard. Just ignore for now that the sign is changing.. You have 3/2=3/2^1, then 3/4=3/2^2. then 3/8=3/2^3. So the nth term will be 3/2^n. How do you get the sign to change?

7. Originally Posted by Johnny_L
(3/2)(-1/2)^(n-1)

3 X-1 = -3; 2 X 2 =4

3/4^(n-1)

why aren't we multiplying numerator by numerator and denominator by denominator?

I don't understand how we get to this

a(n) = 3*(-1)^(n-1)/2^(n)

form at all...
stapel showed you how but possibly this is more clear
(3/2)(-1/2)^(n-1)
=$\frac{3}{2} * (-\frac{1}{2})^{n-1}$

=$3 * \frac{1}{2} * (-\frac{1}{2})^{n-1}$
=$3 * \frac{1}{2} * (-1)^{n-1} * (\frac{1}{2})^{n-1}$
=$3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}$
=$3 * (-1)^{n-1} * (\frac{1}{2})^{n}$
=$3 * (-1)^{n-1} * (\frac{1}{2^{n}})$
= 3 (-1)^(n-1)/2^n

8. Originally Posted by Ishuda
= 3 (-1)^(n-1)/2^n
Agree...however, 3(-1)^(n+1) / 2^n is also correct.

Seems real strange that "the book" gives that
"complicated looking" solution.
Why not simply leave it at:
a[r^(n-1)] where a = 3/2 and r = -1/2

9. Originally Posted by Ishuda
stapel showed you how but possibly this is more clear
(3/2)(-1/2)^(n-1)
=$\frac{3}{2} * (-\frac{1}{2})^{n-1}$

=$3 * \frac{1}{2} * (-\frac{1}{2})^{n-1}$
=$3 * \frac{1}{2} * (-1)^{n-1} * (\frac{1}{2})^{n-1}$
=$3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}$
=$3 * (-1)^{n-1} * (\frac{1}{2})^{n}$
=$3 * (-1)^{n-1} * (\frac{1}{2^{n}})$
= 3 (-1)^(n-1)/2^n
ok, im starting to get it but how do we get from here

=$3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}$
to here
=$3 * (-1)^{n-1} * (\frac{1}{2})^{n}$

One of the 1/2 disappears and the n-1 turns to n. I don't understand

10. Originally Posted by Johnny_L
ok, im starting to get it but how do we get from here

=$3 * (-1)^{n-1} * \frac{1}{2} * (\frac{1}{2})^{n-1}$
to here
=$3 * (-1)^{n-1} * (\frac{1}{2})^{n}$

One of the 1/2 disappears and the n-1 turns to n. I don't understand
As pointed out earlier: What do you get when you combine the (1/2)^1 and the (1/2)^(n-1)?

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