1. ## geometric sequences

Given the following geometric sequence: 32, ____, ____,4,.........

Find the missing two terms

find the nth term

find the sum of the first 5 terms

2. Originally Posted by Airjunkie
Given the following geometric sequence: 32, ____, ____,4,.........

Find the missing two terms

find the nth term

find the sum of the first 5 terms

Examine $\displaystyle a_n=32\bigg(\frac{1}{2}\bigg)^{n}$

3. so i solve that by plugging in 2, 3, and 5 in for n? That would give me the first 5 numbers.

4. Originally Posted by Airjunkie
so i solve that by plugging in 2, 3, and 5 in for n? That would give me the first 5 numbers.
Yes...the corresponding natrual number you put in will yield the n-th term

5. I have no idea what im doing.

6. Originally Posted by Airjunkie
I have no idea what im doing.
Explain what you would do? Give me any inkling you have?

7. I dont know,

I thought i could do

Gn = G1 * r^n-1

8. Originally Posted by Airjunkie
I dont know,

I thought i could do

Gn = G1 * r^n-1

Do you agree that this is the correct sequence? and yes that is the value for the sum of a geometric series

9. No answers eh? Well shoot, if anyone can help me ill try to check this later tonight, maybe i can sneak into the library b4 math.

Thanks,

Airjunkie

10. Originally Posted by Airjunkie
Given the following geometric sequence: 32, ____, ____,4,.........

Find the missing two terms

find the nth term

find the sum of the first 5 terms

Ok based on the given we can observe that the sequence is decreasing by n(1/2). That gives us the complete sequence: 32,16,8,4.
There is a formula for this but it requires two consecutive numbers in a series which is rn=an / an-1

the nth term can be calculated by the formula an=a1r^n-1. We already determined that r=1/2 above. So we just need to plug in our values
an=32* (1/2)^n-1. We can now use this formula to calculate any of the n terms we would like.

Sum of the first 5 terms is done by the formula: Sn=a1 (1-r^n)/1-r
So we plug in our Values:
S5=32(1-1/2^5)/1-1/2 >> S5=32(1-1/32)/.5 >> S5=32(31/32)/.5 S5 = 62

For the last part we use the formula: Sinfinity= a1 /1-r
this formula has a condition though the conditions are "r" must be between -1 & 1 for it to have an infinite series. Since our "r" value was 1/2 we are able to use this.

Sinfinity= 32/1-.5 = 32/.5 = 64

Hope this helps