Question 460555
Let x-3, x-2, ..., x+3 be the seven integers, with


*[tex \LARGE \frac{(x-3)^2 + (x-2)^2 + ... + (x+3)^2}{7} = 53]


*[tex \LARGE (x-3)^2 + (x-2)^2 + ... + (x+3)^2 = 371]


If we were to expand this all out, we would see that all the x terms would disappear (-6x on the left term cancels out 6x on the right, and so on). We are left with


*[tex \LARGE (x^2 + 9) + (x^2 + 4) + (x^2 + 1) + x^2 + (x^2 + 1) + (x^2 + 4) + (x^2 + 9) = 371]


*[tex \LARGE 7x^2 + 28 = 371]


*[tex \LARGE 7x^2 = 343]


*[tex \LARGE x^2 = 49 \Rightarrow x = \pm 7].


The average of the integers x-3, x-2, ..., x+3 is


*[tex \LARGE \frac{(x-3) + (x-2) + ... + (x+3)}{7}] Again, the constant terms cancel, leaving


*[tex \LARGE \frac{7x}{7} = x]


So the average of the original seven integers is either 7 or -7.