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The sum of three numbers in A.P. is 12 and the sum of their cube is 288. Find the numbers. |
Answer» 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 a 3 – 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 = \(\sqrt{4}\) 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 |
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