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in Arithmetic Progression by (31.3k points)
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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.

1 Answer

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Best 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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