All of your answers for the first problem are right, even D.
For problem 1D I crunched some #'s and found that if you plot the points of the averages, you get a linear graph, and the function is avg=2x+2, or y=2x+2. So you want the avg of the sum of 999th # in the series, just plug in 2(999)+2, 1998+2=<u>2000</u>. Try it. If you want the sum of the 5th number and all the numbers in the series before it... let's try it out. The numbers on the left represent x, they just indicates which number in the series we're at. The number in the middle column will be the actual numbers in the series, and the numbers on the right column are the sum so far:
<pre>
x|#in series|sum so far
1| 4 | 4
2| 8 | 12
3| 12 | 24
4| 16 | 40
5| 20 | 60
</pre>
For the 5th #, the sum is 60. So avg is 60/5=<font color=red>12</font>.
Of course, it would take forever to make a table that goes all the way up to 999,
but first let's prove that avg=2x+2 works.
x=5, so avg=2*5+2=10+2=<font color=red>12</font>. Voila! Try it with any number you want.
So again, x=999. avg=2*999+2=1998+2=2000. ^^ Hope this helps.
If you make a similar table for 1B and 1C, you can use avg=x+1.
Taking this a bit further, I think whenever you have a series like these, say you start with 3 and go up in increments of 3, like 3, 6, 9, 12... and you want the average of the sums, you just take the interval, in this case 3, and divide it by 2. That's 1.5. Then y=1.5x+1.5, just in case you need to do something like this again with a different interval.