The sum of three numbers in A.P. is 12 and the sum of their cube is 288. Find the numbers.
Let the numbers be (a – d), a, (a+ d)
A – d + a + a+ d = 12
3a = 12
a = 4
According to question,
(a – d)3 + a3 + (a + d)3 = 288
a3 – d3 – 3ad (a – d) + a3 + a3 + d3 + 3ad (a + d) = 288
3a3 – 3a2d + 3ad2 + 3a2d + 3ad2 = 288
3(4)3 + 6(4) d2 = 288
24d2 = 96
d2 = 4
d = �4
Therefore, when a = 4 and d = 2
a – d = 4 – 2 = 2
a + d = 4 + 2 = 6
When a = 4 and d = -2
a – d = 4 + 2 = 6
a + d = 4 – 2 = 2