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

Find the missing two terms

find the nth term

find the sum of the first 5 terms

find the sum of all the terms. Explain your answer

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- May 1st 2008, 04:59 PMAirjunkiegeometric 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

find the sum of all the terms. Explain your answer - May 1st 2008, 05:12 PMMathstud28
- May 1st 2008, 05:15 PMAirjunkie
so i solve that by plugging in 2, 3, and 5 in for n? That would give me the first 5 numbers.

- May 1st 2008, 05:18 PMMathstud28
- May 1st 2008, 05:21 PMAirjunkie
I have no idea what im doing.

- May 1st 2008, 05:26 PMMathstud28
- May 1st 2008, 05:37 PMAirjunkie
I dont know,

I thought i could do

Gn = G1 * r^n-1

I need some help bad. - May 1st 2008, 05:42 PMMathstud28
- May 1st 2008, 08:49 PMAirjunkie
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 - May 1st 2008, 11:03 PMDaddycakes
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