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