Question 1161837
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Three consecutive odd integers form an arithmetic sequence.<br>
In an arithmetic sequence of three terms, the middle term is the average of the first and third -- i.e., the sum of the first and third is two times the middle term.<br>
So if the sum of the first and third is 13 more than the middle, the middle term must be 13.<br>
And so the three odd integers are 11, 13, and 15.<br>
Putting the above argument in terms of formal algebra....<br>
Let the three integers be x-2, x, and x+2<br>
The sum of the first and third is 13 more than the middle:<br>
{{{(x-2)+(x+2) = (x)+13}}}
{{{2x = x+13}}}
{{{x = 13}}}<br>
And the three integers are
x-2 = 11
x = 13
x+2 = 15<br>