Question

If the square of sum of three positive consecutive natural numbers exceeds the sum of their squares by 292, then what is the largest of the three numbers?

Answer 1

a-1,a& a+1 are three consecutive numbers. Use the above equation

Answer 2

The largest of the three numbers is 8.

• Let the three positive consecutive natural numbers be x-1 , x , x+1.

• Now the square of the sum of these numbers exceeds the sum of their squares by 292.

• Now,

(x-1 + x + x+1)² = (x-1)² + x² + (x+1)² + 292

(3x)² = x²-2x+1 + x² + x²+2x+1 + 292

9x² = 3x² + 294

6x² = 294

x² = 49

x = 7 (As the number is a natural number)

• Now, the largest number will be x+1 = 7+1 = 8
• Largest number among the three numbers is 8.