Question 460555
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Let *[tex \Large x] represent the first integer.  Then the next one must be *[tex \Large x\ + 1], the one after that is *[tex \Large x\ +\ 2], and so on.


So the first integer squared is *[tex \Large x^2]


The second integer squared is *[tex \Large (x\ +\ 1)^2\ =\ x^2\ +\ 2x\ +\ 1]


The third one is *[tex \Large (x\ +\ 2)^2\ =\ x^2\ +\ 4x\ +\ 4]


And so on.


Multiply out the squared binomials for the seven integers, *[tex \Large x] through *[tex \Large x\ +\ 6], then add the seven expressions, collecting like terms.  Divide your result by 7 to to get a quadratic trinomial representing the average of the 7 squared integers.  Set this trinomial equal to 53.


Put your quadratic into standard form and solve.  You will obtain two roots, each of which can be the first of your series of 7 integers.


You can also add *[tex \Large x] through *[tex \Large x\ +\ 6] and then divide by 7 to get a binomial expression for the average of the 7 integers.  Once you have solved the quadratic for *[tex \Large x], you can substitute into this binomial expression to get the value of the average of the integers.  Remember to do it for both roots.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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