Question

addition of square of three consecutive natural numbers is 110. Then in this find out the smallest natural number.​

Answer 1

Answer:

5, 6, and 7

Step-by-step explanation:

Let three consecutive natural numbers be x, x + 1 and x +2.

Then according to problem

(x)2 + (x + 1)2 + (x + 2)2 = 110

⇒ x2 + x2 + 1 + 2x + x2 + 4 + 4x – 110 = 0

⇒  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 + 7 = 0 or x – 5 = 0

⇒ x  = -7 or  x = 5

But x = - 7 is rejected as it is not a natural number.

Then x = 5

Hence,  required numbers are 5, (5 + 1), (5 + 2)

i.e.,  5, 6  and 7.

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. hope it helps.......

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.