Here's the cool thing:  We're NOT going to have to keep making these tables!  Since all arithmetic sequences behave the same way (change by adding a number), we can get a formula that works for ALL of them!

Check out the pieces of our last guy:

       an = 3 + ( n - 1 )( 4 ) ... the nth term is an ... a1 is 3 ... 3 is d = difference

Let's check to see if this thing works on another arithmetic sequence:

How about this guy?

6 , 8 , 10 , 12 , 14 , ...

This would be the formula:

an = 6 + ( n - 1 )( 2 ) ... a1 is 6 ... 2 is d

Let's use it to finda10:

a10 = 6 + 9( 2 ) = 24 ... a10 is n ... 9 is n - 1

Now, let's check it the long way:

6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , ... a10 is 24
Yep!

The cool thing is that you can use this formula to find the 2000th term:

a2000 = 6 + 1999( 2 ) = 4004