# Thread: Geometric series #2

1. ## Geometric series #2

Another one is this one that I am unsure how to go about it:

Find the nth term of the geometric series: 1024, 512, 256, 128,...

2. Originally Posted by mike1
Another one is this one that I am unsure how to go about it:

Find the nth term of the geometric series: 1024, 512, 256, 128,...
an = a1 *r^(n-1) --------------***

1024, 512, 256, 128,...
So, a1 = 1024.

r = 512/1024 = 1/2
r = 256/512 = 1/2
r = 128/256 = 1/2

Hence, nth term is
an = a1 *r^(n-1)
an = 1024 *(1/2)^(n-1)
an = 1024 *[(1/2)^n / (1/2)]
an = 1024 *[2 *(1/2)^n]
an = 2048 *[1^n / 2^n]
an = 2048 *[1 / 2^n]
an = 2048 / 2^n ----------------answer.

Check,
a3 = 256 --------given
256 =? 2048 / (2^3)
256 =? 2048 / 8
256 =? 256
Yes, so, OK.

3. Hello, Mike!

Find the nth term of the geometric series: 1024, 512, 256, 128, ...
If you have the formula and can recognize the parts, you're all set.

The nth term of a geometric series: a_n .= .ar^{n-1}

We can see that the first term is: a = 1024, right?

And we should be able figure out that: r = 1/2

Then we plug them into the formula . . .

So where exactly is your difficulty?

4. My problem is that I am doubting myself. It turns out easier than I make it out to be. You are a great help.