SOLUTION: The average of the squares of seven consecutives integers is 53. The average of these integers is?

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Question 460555: The average of the squares of seven consecutives integers is 53. The average of these integers is?
Found 3 solutions by solver91311, Edwin McCravy, richard1234:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Let represent the first integer. Then the next one must be , the one after that is , and so on.

So the first integer squared is

The second integer squared is

The third one is

And so on.

Multiply out the squared binomials for the seven integers, through , 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 through and then divide by 7 to get a binomial expression for the average of the 7 integers. Once you have solved the quadratic for , 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

My calculator said it, I believe it, that settles it
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Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
The average of the squares of seven consecutives integers is 53. The average of these integers is?

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
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!
Let x-3, x-2, ..., x+3 be the seven integers, with





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







.

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

Again, the constant terms cancel, leaving



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