Anonymous
Anonymous asked in Education & ReferenceHomework Help · 9 years ago

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

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

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  • 9 years ago
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    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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