Question

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

Answer 1

1,4,7

Step-by-step explanation:

the sum of three no of an A.P are given below

$(a-d),a,(a+d)\\a-d+a+a+d=12\\3a=12\\a=4$

the sum of their cube are follows

$(a-d)^{3},a^{3},(a+d)^{3}\\(a-d)^{3}+a^{3}+(a+d)^{3}=288\\using(a-b)^{3}or(a+b)^{3}\\a^3-b^3-3ab(a-b)+a^3+b^3+3ab(a+b)\\a^3-d^3-3ad(a-d)+a^3+a^3+d^3+3ad(a+d)\\a^3-d^3-3a^2d+3ad^2+a^3+a^3+d^3+3a^2d+3ad^2\\3a^3+6ad^2=288\\a^3+2ad^2=136-(i)$

put  $a=4$  in equation (i)

$4^3+2\times 4\times d^2=136$

$64+8d^2=136\\8d^2=136-64\\8d^2=72\\d^2=\frac{72}{8} \\d^2=9$

$d=\sqrt{9}$

$d=3$

number are follow

$a-d,a,a+d\\4-3,4,4+3\\1,4,7$