Question

The sum of squares if three consecutive natural numbers is 110

Answer 1

i think there is an error in the questions

Answer 2

Let:-

The three consecutive natural numbers are x, x + 1 and x + 2

Then:-

According to Question

➨ (x)² + (x + 1)² + (x + 2)² = 110

➨ x² + (x² + 1² + 2x) + (x² +3² + 2*2*x) =110

➨ x² + x² + 1² + 2x + x² +3² + 2*2*x =110

➨ 3x² + 1 + 4 + 2x + 4x = 110

➨ 3x² + 5 + 6x = 110

➨ 3x² + 6x + 5 - 110 = 0

➨ 3x² + 6x - 105 = 0

Now divide the equation by 3

➨ x² + 2x - 35 = 0

➨ x² + 7x -5x - 35 = 0

➨ x(x + 7) -5(x + 7) = 0

➨ (x + 7) (x - 5) = 0

Therefore:-

➨ x + 7 = 0

➨ x = -7 [Not valid] and

➨ x - 5 = 0

➨ x = 5

Hence:-

The required smallest natural number is 5.