The sum of 3 numbers in an arithmetic progression is 12 and the sum of their square is 66 find the numbers
Let the three numbers be:
n1 = a
n2 = a + d (where d is the common difference between terms in the arithmetic progression)
n3 = a + 2d
Then,
n1 + n2 + n3 = 12
n1^2 + n2^2 + n3^2 = 66
Substituting values for the three numbers,
a + (a + d) + (a + 2d) = 12
a^2 + (a + d)^2 + (a + 2d)^2 = 12
Expanding the brackets,
3a + 3d = 12
3a^2 + 6ad + 5d^2 = 66
Using a = 4 - d, from 3a + 3d = 12,
3(4 – d)^2 + 6(4 – d)d + 5d^2 = 66
3(16 – 8d + d^2) + 24d – 6d^2 + 5d^2 = 66
48 – 24d + 3d^2 + 24d - d^2 = 66
2d^2 = 18
d = 3
a = 4 – 3 = 1
Then the three numbers are:
n1 = 1, n2 = 4, n3 = 7