Q2 of 176 Page 8

The sum of three numbers in A.P. is 12 and the sum of their cubes is 408. Find the numbers.

Let the three numbers are in AP = a, a + d, a + 2d


According to the question,


The sum of three terms = 12


⇒ a + (a + d) + (a + 2d) = 12


⇒ 3a + 3d = 12


⇒ a + d = 4


⇒ a = 4 – d …(i)


and the sum of their cubes = 408


⇒ a3 + (a + d)3 + (a + 2d)3 = 408


⇒ (4 – d)3 + (4)3 + ( 4 – d + 2d)3 = 408 [from(i)]


⇒ (4 – d)3 + (4)3 + ( 4 + d)3 = 408


⇒ 64 – d3 + 12d2 – 48d + 64 + 64 + d3 + 12d2 + 48d = 408


⇒ 192 + 24d2 = 408


⇒ 24d2 = 408 – 192


⇒ 24d2 = 216


⇒ d2 = 9


⇒ d = √9


⇒ d = ±3


Now, if d = 3, then a = 4 – 3 = 1


and if d = – 3, then a = 4 – ( – 3) = 4 + 3 = 7


So, the numbers are →


if a = 1 and d = 3


1, 4, 7


and if a = 7 and d = – 3


7, 4, 1


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