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If the sum of the squares three consecutive natural numbers is 110, then the smallest of these natural numbers is = ?
Answer & Solution
Correct Answer:
Option
D
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
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
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