Question

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

Answer 1

\[\text { Let the numbers be } (a - d), a, (a + d) . \]

\[\text { Sum } = a - d + a + a + d = 12\]

\[ \Rightarrow 3a = 12\]

\[ \Rightarrow a = 4\]

\[\text { Also }, (a - d )^3 + a^3 + (a + d )^3 = 288\]

\[ \Rightarrow a^3 - d^3 - 3 a^2 d + 3a d^2 + a^3 + a^3 + d^3 + 3 a^2 d + 3a d^2 = 288\]

\[ \Rightarrow 3 a^3 + 6a d^2 = 288\]

\[ \Rightarrow 3 \left( 4 \right)^3 + 6 \times 4 \times d^2 = 288\]

\[ \Rightarrow 192 + 24 d^2 = 288\]

\[ \Rightarrow 24 d^2 = 96\]

\[ \Rightarrow d^2 = 4\]

\[ \Rightarrow d = \pm 2\]

\[\text { When a = 4, d = 2, the numbers are } 2, 4, 6 . \]

\[\text {  When a = 4, d = - 2, the numbers are } 6, 4, 2 .\]