If the sum of the squares three consecutive natural numbers is 110, then the smallest of these natural numbers is = ?

Solution (By Examveda Team)

Let the three consecutive numbers are x, x + 1, x + 2
According to question
x² + (x + 1)² + (x + 2)² = 110
⇒ x² + x² + 1 + 2x + x² + 4 + 4x = 110
⇒ 3x² + 6x + 5 = 110
⇒ 3x² + 6x - 105 = 0
⇒ x² + 2x - 35 = 0
⇒ x² + 7x - 5x - 35 = 0
⇒ x(x + 7) -5(x + 7) = 0
⇒ (x + 7)(x - 5) = 0
⇒ x = 5 & -7
[∴ (-) value can not considered]
∴ Smallest number is = 5