Lesson A shortcut to finding the sum of 2x+1 consecutive integers

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This Lesson (A shortcut to finding the sum of 2x+1 consecutive integers) was created by by jgr45(22) About Me : View Source, Show
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Whenever a consecutive-integer problem asks for 3 or 5 or 7 consecutive integers that add up to a particular sum, most people like to set x (or whatever variable) equal to the smallest number -- and that's fine. However, I prefer to set x equal to the middle number and here you'll see why:
For example, let's say we're instructed to find five consecutive odd integers that total 545. If we let x equal the MIDDLE number, then this is what we would have:
first number = x-4
second number = x-2
third number = x
fourth number = x+2
fifth number = x+4
(x-4)+(x-2)+x+(x+2)+(x+4) = 545
Now, the associative property of addition states that we can place sets of parentheses and eliminate sets of parentheses anywhere we like in an addition problem without affecting the sum.
The commutative property of addition states that we can add numbers in any order we like and the sum also will not be affected.
Keeping these rules in mind, look at what happens:
(x+x+x+x+x)+(-2+2)+(-4+4) = 545
5x = 545 (WHAT??? That's right, the 2's and 4's are eliminated!) :)
x = 109
Remember, we said this was the middle number, so our set of numbers are 105, 107, 109, 111, and 113.
Had the problem said simply integers and not odd integers, the solution would have been 107, 108, 109, 110, and 111.
In fact, problems like this can easily be done mentally. Simply set your variable multiplied by the number of numbers (if odd) you are asked to find and set it equal to the sum you are given, and when solving, set your variable equal to the middle number. One more example:
Find three consecutive (even) integers that total 162.
3x = 162 (bypassing the first step, for reason above)
x = 54 (middle number)
Numbers are 53, 54, and 55. Or if they had to be even numbers, 52, 54, and 56.
That quick and easy. :)

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