Sum of squares of three consecutive natural numbers is 110. Find the number?
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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.
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