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