Question -
Answer -
Given:
The sum of threenumbers is 12
Let us assume thenumbers in AP are a – d, a, a + d
So,
3a = 12
a = 4
It is also given thatthe sum of their cube is 288
(a – d)3 +a3 + (a + d)3 = 288
a3 – d3 –3ad(a – d) + a3 + a3 + d3 +3ad(a + d) = 288
Substitute the valueof a = 4, we get
64 – d3 –12d(4 – d) + 64 + 64 + d3 + 12d(4 + d) = 288
192 + 24d2 =288
d = 2 or d = – 2
Hence, the numbers area – d, a, a + d which is 2, 4, 6 or 6, 4, 2