# The sum of three numbers in A.P. is 12, and the sum of their cubes is 288. Find the numbers.

assume the numbers in AP are a - d, a, a + d

Given that the sum of three numbers is 12

To find: the first three terms of AP

So,

3a = 12

a = 4

It is also given that the 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

substituting 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 are a - d, a, a + d which is 2, 4, 6 or 6, 4, 2

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