If the square of sum of three positive consecutive natural numbers exc...
Given Information:
- Let the three consecutive natural numbers be x, x+1, and x+2.
Given Condition:
- The square of the sum of the three numbers exceeds the sum of their squares by 484.
Formulas:
- Sum of the three numbers: x + x+1 + x+2 = 3x + 3
- Square of the sum of the three numbers: (3x + 3)^2
- Sum of their squares: x^2 + (x+1)^2 + (x+2)^2
Equation:
- (3x + 3)^2 = x^2 + (x+1)^2 + (x+2)^2 + 484
Solve the Equation:
- Expand the left side: 9x^2 + 18x + 9 = x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 + 484
- Simplify: 9x^2 + 18x + 9 = 3x^2 + 6x + 489
- Rearrange: 6x^2 + 12x - 480 = 0
- Divide by 6: x^2 + 2x - 80 = 0
- Factorize: (x + 10)(x - 8) = 0
- The possible values for x are -10 and 8. Since the numbers are positive, x = 8.
Calculate the Largest Number:
- Largest number = x + 2 = 8 + 2 = 10
Therefore, the largest of the three numbers is 10. Hence, the correct answer is option 'D'.