Help me find the average of the 1000 whole numbers from 1 to 1000,inclusive plz,i really need help?

Question

I dont know how to do this, because it will take to long to add up 100 numbers and divide it by 1000. So i need help!!!!!!!!!!!!

Answer 1

Sum from 1 to n = [(n)(n+1)]/2 (memorize this)
Average = sum / n

So average from 1 to n
= [[(n)(n+1)]/2]/n
= [(n)(n+1)]/2n

n = 1000
= (1000*1001) / 2000
= 500.5

Answer 2

you have 500 pairs of numbers that add up to 1001 (1+1000 , 2+999 through 500+501) the sum of all of the numbers is 500 times 1001 or 500,500. the average of each pair is 500.5, which is also the average of the entire series.

an average is the sum of all of the numbers divided by the number of numbers you are averaging. a = s/n
500,500/1000 = 500.5

Answer 3

Sum = n*(a+b) / 2
Average = sum / n = n*[(a+b) / 2] / n = (a+b) / 2

** n: total numbers, a: first number, b: last number

Remember, the "average" is always the first number plus the last number, and divided by 2. It's that simple!

(1+1000) / 2 = 500.5

Answer 4

look at pairs of numbers:
1 + 1000 = 1001
2 + 999 = 1001
3 + 998 = 1001
500 + 501 = 1001

So you have 500 pairs that each sum to 1001
The average then is 500(1001)/1000 = 500.5

Answer 5

1+2+3+4+...+998+999+1000
=（1+999）+（2+998）+...+（499+501）+500+1000
=1000+1000+...+1000+500+1000
=499*1000+500+1000
=499000+1500
=500500

500500/1000=500.5


just for reference

Answer 6

500.5