Question

The sum of the squares of five consecutive natural numbers is 1455. Find the numbers.

Answer 1

Let the five consecutive integers be n, n + 1, n + 2, n + 3, n + 4 then,

n² + (n + 1)² + (n + 2)² + (n + 3)² + (n + 4)² = 1455

n² + n² + 1+ 2n + n² + 4n + 4 + n² + 6n + 9 + n² + 8n + 16 = 1455

[(a+b)² = a² + 2ab + b²]

5n² + 20n + 30 = 1455

5n² + 20n – 1425 = 0

5n² + 20n – 1425 = 0

n² + 4n – 285 = 0

∴ Compare with ax² + 2bx + c = 0

`a = 1, b = 4, c = - 285`

b² − 4ac = 4² − 4 × 1 × (−285)

= 16 + 1140 = 1156

n = `(-b ± sqrt(b^2- 4ac))/(2a)`

= `(-4 ± sqrt1156)/2`

= `(-4 ± 34)/2`

= `(-4 ± 34)/2` or `(-4 -34)/2`

`n = 15n -19`

As, `n = - 19` is not natural
Hence, the five consecutive numbers are 15,16, 17, 18, 19