Find three consecutive positive integers such that the product of the larger two is equal to twice the second integer plus twice the sum of all three integers.

Answer 1

Let the three consecutive positive integers be x-1, x and x+1

The product of the two larger integers is x(x+1) = x² + x

Twice the 2nd integer is 2x

The sum of the three integers is 3x

So twice the sum of the three integers is 2*3x=6x

So the sum of twice the 2nd integer and twice the sum of the three integers is 3x + 6x =9x

Now x² + x = 9x. So x² = 8x and x=8.

The the three positives are 7, 8 and 9