Question

Find three consecutive positive numbers such that the square of the middle number exceeds the difference of the squares of the other two by 60.

Answer 1

Let the three consecutive positive numbers be

(n − 1), n, (n + 1)

(n + 1)² − (n − 1)²

= (n² + 2n + 1) − (n² − 2n + 1)

“The square of the middle number exceeds this difference by 60” means:

n² = 4n + 60

n² − 4n − 60 = 0

(n − 10) (n + 6) = 0

n = 10 or n = −6

Since the numbers must be positive, we take n = 10

9, 10, 11​