Question 460555
The average of the squares of seven consecutives integers is 53. The average of these integers  is?
<pre>

The tutor above, who solved it before me, suggested doing it a
harder way, by using

x,  x+1, ... , x+6.

But it's much easier to let x be the middle integer instead of
the smallest integer.  I see that the tutor below, who solved it
after I did, solved it the same way I did.

There are two solutions:  7 and -7.  Here's how to 
find them:

Let the 4th integer be x, then the 7 integers are

x-3, x-2, x-1, x, x+1, x+2, x+3

and their average is their sum over 7 

x-3+x-2+x-1+x+x+1+x+2+x+3
—————————————————————————
            7

Collecting terms:

          7x
          ——
           7

           x


Their squares are:

(x-3)², (x-2)², (x-1)², x², (x+1)², (x+2)², (x+3)²

or

x²-6x+9, x²-4x+4, x²-2x+1, x², x²+x+1, x²+4x+4, x²+6x+9

The average of these squares is the sum of those squares
over 7

x²-6x+9+x²-4x+4+x²-2x+1+x²+x²+x+1+x²+4x+4+x²+6x+9
—————————————————————————————————————————————————
                       7

Collecting terms:

7x²+28
——————
   7

Setting that equal to 53

7x²+28
—————— = 53
   7

Multiply through by 7

7x²+28 = 371

   7x² = 343
    
    x² = 49

     x = ±7

And since x is their average. the answer is ±7

So there are two solutions.

The integers are then either

-10, -9, -8, -7 , -6, -5, -4

           or

  4, 5, 6, 7, 8, 9, 10 

Checking: the average of the square of either are

  4² + 5² + 6² + 7² + 8² + 9² + 10²
  —————————————————————————————————
               7

  16 + 25 + 36 + 49 + 64 + 81 + 100
  —————————————————————————————————
                 7

               371
               ———
                7

               53

So we're correct.  There are two solutions.
Perhaps you were told only to find the positive 
consecutive integers. If not, give both solutions.

Edwin</pre>