Question
the sum of the reciprocals of 3 consecutive positive integers is 47 divided by the product of the integers. what is the smallest of the 3 integers
Answers
1/(x-1) + 1/x + 1/(x+1) = 47/((x-1)x(x+1))
x(x+1) + (x-1)(x+1) + x(x-1) = 47
x^2 + x + x^2 - 1 + x^2 - x
3x^2 - 1 = 47
3x^2 = 48
x^2 = 16
x = 4
so, the numbers are 3,4,5
x(x+1) + (x-1)(x+1) + x(x-1) = 47
x^2 + x + x^2 - 1 + x^2 - x
3x^2 - 1 = 47
3x^2 = 48
x^2 = 16
x = 4
so, the numbers are 3,4,5
Wrong
Related Questions
The sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the prod...
The sum of the reciprocals of two consecutive positive integers is 17/12. Write an equation that can...
2. The sum of the reciprocals of two consecutive positive integers is 17/72. Write an equation that...