Answer :

It is given in the question that, sum of three A.P. = 12

Also, Sum of cubes of three A.P. = 288

Let us assume the three numbers in A.P. are a – d, a and a + d

∴ a – d + a + a + d = 12

3a = 12

Also, (a – d)^{3} + (a)^{3} + (a + d)^{3} = 288

a^{3} – 3a^{2}d + 3ad^{2} – d^{3} + a^{3} + a^{3} + 3a^{2}d + 3ad^{2} + d^{3} = 288

3a (a^{2} + 2d^{2}) = 288

3 × 4 (4 × 4 + 2 × d × d) = 288

2d^{2} = 8

∴ d^{2} = 4

Now, when the value of a = 4 and d = 2 then the numbers will be:

a – d = 4 – 2 = 2

a = 4

a + d = 4 + 2 = 6

Or, when a = 4 and d = - 2 then the numbers will be:

a – d = 4 – (- 2) = 6

a = 4

a + d = 4 + (- 2) = 2

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