Question

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

Answer 1

Let the numbers be (a - d), a, (a + d)
Sum = a - d + a + a + d = 12
$\Rightarrow 3a=12\Rightarrow a=4$
Also, $(a-d{)}^3+a^3+(a+d{)}^3=288$
$\Rightarrow a^3-d^3-3a^{2}d+3a{d}^2+a^3+a^3+d^3+3a^{2}d+3a{d}^2=288\Rightarrow 3a^3+6a{d}^2=288\Rightarrow 3(4{)}^3+6\times 4\times d^2=288\Rightarrow 192+24d^2=288\Rightarrow 24d^2=96\Rightarrow d^2=4\Rightarrow d=\pm 2$
When a = 4, d = 2, the numbers are 2, 4, 6
When a = 4, d = -2, the numbers are 6, 4, 2