<p>Do any of you know a quick way to calculate consecutive EVEN or ODD integers?? Because I know that to calculate consecutive integers you do: biggest int - smallest int + 1... but I often see questions about consecutive even or odd integers on the practice tests I take.</p>
<p>Thanks!</p>
<p>Bump* I could use this also! Thank you! :)</p>
<p>Add or subtract 2 instead of one.</p>
<p>@Yankee Belle: I tried but it doesn’t work…</p>
<p>Here is an example. The sum of 5 consecutive odd integers = 165. What is the value of the middle number.<br>
N
N +2
N +4
N +6</p>
<h2>N +8</h2>
<p>5N+20 = 165
-20 -20
5N = 145
N = 29
The middle value = N + 4 = 29+4</p>
<p>Does that help?</p>
<p>Let’s say you have n consecutive odd integers. As examples, let’s have one set A={1,3,5,7} and B={3,5,7,9,11}. So, for A, n=4; for B, n=5. Notice how 1+7=3+5, and 7=(5+7)/2=(3+11)/2. Therefore, we can come up with this formula: SUM=(greatest + least)*n/2. Plug these numbers in and you’ll see that this formula works when n is both even and odd.</p>
<p>Now that you have the sum, the average should be easy:AVG=SUM/n=(greatest + least)/2. Same thing applies for even numbers.</p>
<p>use the ‘‘sum of arithmetic progression’’ this only applys when the difference in numbers are constant.</p>
<p>Formula = n/2(2a+(n-1)d) </p>
<p>so n= number of terms
a= the first number
d= difference in numbers</p>
<p>i think this is the easiest way and hope this helps</p>
<p>Let Odd=p, Even=q</p>
<p>p=1 (mod 2)
q=0 (mod 2)</p>
<p>p=2k+1, q=2c</p>
<p>Consecutive odds: (2k+1),(2k+3),(2k+5)
Consecutive even: (2k),(2k+2),(2k+4)</p>