# 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 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...

4. 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?

5. 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

6. 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

7. 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)?

8. 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
When you multiply two factors with the same base you keep the base and add the exponents. Just need to know that 1/2 = (1/2)^1 and that 1+ (n-1) is n.

9. Originally Posted by Jomo
When you multiply two factors with the same base you keep the base and add the exponents. Just need to know that 1/2 = (1/2)^1 and that 1+ (n-1) is n.
mind = blown

i've only done that when the variables were bases.

but it works with regular numbers too. makes sense.

thanks!

10. Originally Posted by Johnny_L
mind = blown

i've only done that when the variables were bases.

but it works with regular numbers too. makes sense.

thanks!
Variables are place holders for numbers. So yeah, if something works for variables it must work for numbers!

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•