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# Find the average of the 3000 counting numbers from 1 to 3000, inclusive.?

Please explain how you did this. Step by step. Thanks!

### 2 Answers

- 9 years agoFavorite Answer
I'm showing you how to derive the equation that you use instead of just plugging it into the end equation and not understanding how it works. I'll do the average for 1 to 10 just for an example. You can do the same thing for 1 to 3000 or any other number.

Basically, look at it this way:

1,10 =11

2,9=11

3,8=11

4,7=11

5,6=11

If you match up the pairs starting from either end for x terms, then you get x/2 pairs of the sum of the first number and the last number (which equals x+1 if you start at 1 and end at x). You multiply the sum of the first and last terms (x+1) by the number of pairs (x/2) in order to get the sum of all the numbers from 1 to x, then divide that by x to get the average.

As an expression, that would be [(x+1)(x/2)]/x

Simplified would be (x+1)/2

For 3000, the answer would be 1500.5

Note: This expression will work for an odd number of terms, and only if you start from 1. If you were to start from x and end at y, you should use:

(x+y)[(y-x+1)/2]/(y-x+1)

Simplified, that would be (x+y)/2

Here, y-x+1 is an expression for the number of terms. If you go from 2 to 5 there are four terms, so 5-2+1.

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- 9 years ago
Step 1 : Find the sum of all counting numbers from 1 to 3000 using the formula

( L² - F² + F + L)/2 where L is the last number and F is the first number

Sum = (3000^2 - 1 + 3000 + 1)/2 = 4501500

Step 2 : Divide by the number of numbers..... i.e, 3000

Average = 4501500 / 3000 = 1500 . 5

There!

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