The sum of three numbers in A.P. is 12 and sum of their cubes are 288. Find the numbers.

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

a³ – 3a²d + 3ad² – d³ + a³ + a³ + 3a²d + 3ad² + d³ = 288

3a (a² + 2d²) = 288

3 × 4 (4 × 4 + 2 × d × d) = 288

_2d² = 8

∴ d² = 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