question_answer

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

Answer:

Let the three numbers in A.P. be \[a-d,\text{ }a,\text{ }a+d\]

Now, \[a-d+a+a+d=12\]

\[3a=12\]

\[\therefore a=4\]

Also, \[{{(4-d)}^{3}}+{{4}^{3}}+{{(4+d)}^{3}}=288\]

\[64-48+12{{d}^{2}}-{{d}^{3}}+64+64+48d+12{{d}^{2}}+{{d}^{3}}=288\]

\[192+24{{d}^{2}}=288\]

\[24{{d}^{2}}=288192\]

\[{{d}^{2}}=\frac{96}{24}=4\]

\[d=\pm 2\]

The numbers are 2, 4, 6 or 6, 4, 2.