find 5 consecutive whole numbers if it is known that the sum of the squares of the first 3 numbers is equal to the sum of the squares of the last 2 numbers.

10, 11, 12, 13, 14

Step-by-step explanation:

Let x represent the first of these 5 consecutive whole numbers.

The next four numbers are x+1, x+2, x+3 and x+4.

Then:

x^2 + (x+1)^2 + (x+2)^2 = (x+3)^2 + (x+4)^2

Expanding this, we get:

x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = (x^2 + 6x + 9 + x^2 + 8x + 16

There are 3 x^2 terms on the left side and 2 x^2 terms on the right side.  Cancel out all of these but for the first x^2 term, on the left, obtaining:

x^2 + 2x + 1 + 4x + 4 = 6x + 9 + 8x + 16.  Combining like terms, we get:

x^2 + 6x + 5              =  14x + 25.

Subtracting 14x + 25 from both sides, we get:

x^2 - 8x - 20 = 0, which factors as follows:

(x - 10)(x + 2) = 0, whose roots are 10 and -2.

Focusing on positive consecutive numbers, we start with x = 10 and continue:  10, 11, 12, 13, 14.

It remains to be shown that another solution is:

-2, -1, 0, 1, 2

the answer is 37

step-by-step explanation:

answer:

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