Question

If the sum of three numbers in AP is 12 and the sum of their cubes is 288, then the numbers are

• A. 2,4,6
• B. 1,4,7
• C. 1,3,5
• D. 3,5,7

Answer 1

The correct option is A 2,4,6

Let the numbers be,

$a-d,a,a+d$

$\therefore a-d+a+a+d=12$

$3a=12$

$\therefore a=4$

now,

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

$(a-d{)}^3+(a+d{)}^2=288-4^3=224$

$\left(2a\right)\left[\right(a-d{)}^2+(a+d{)}^2-(a-d\left)\right(a+d\left)\right]=224$

$(2\times 4)\left[\right(4-d{)}^2+(4+d{)}^2-(4-d\left)\right(4+d\left)\right]=224$

upon solving the above equation, we get,

$16+3d^2=28$

$3d^2=12$

$\therefore d=\pm 2$

Now,

The numbers are,

$2,4,6$ and $6,4,2$