Find the sum of five consecutive natural numbers in which the sum of the squares of the last

Find the sum of five consecutive natural numbers in which the sum of the squares of the last two numbers is equal to the sum of the squares of the first three numbers.

1. Let n1, n2, n3, n4 and n5 be five consecutive numbers, and let n = n3 be the average. Then:

n1 = n – 2;
n2 = n – 1;
n3 = n;
n4 = n + 1;
n5 = n + 2.
2. Let’s compose the equation according to the condition:

n1 ^ 2 + n2 ^ 2 + n3 ^ 2 = n4 ^ 2 + n5 ^ 2;
(n – 2) ^ 2 + (n – 1) ^ 2 + n ^ 2 = (n + 1) ^ 2 + (n + 2) ^ 2;
n ^ 2 – 4n + 4 + n ^ 2 – 2n + 1 + n ^ 2 = n ^ 2 + 2n + 1 + n ^ 2 + 4n + 4;
3n ^ 2 – 6n + 5 = 2n ^ 2 + 6n + 5;
3n ^ 2 – 6n + 5 – 2n ^ 2 – 6n – 5 = 0;
n ^ 2 – 12n = 0;
n (n – 12) = 0;
[n = 0;
[n – 12 = 0;
[n = 0 – not a natural number;
[n = 12.
3. Sum of numbers:

10 + 11 + 12 + 13 + 14 = 5 * 12 = 60.

Answer: 60.