Question

If the sum of three numbers in A.P., is $24$ and their product is $440$, find the numbers.

Answer 1

Let the three numbers in A.P. be $a-d,a,$ and $a+d$
According to given information
$\text{Sum}=(a-d)+\left(a\right)+(a+d)=\mathrm{24...}\left(1\right)\Rightarrow 3a=24\therefore a=8\text{\& Product}=(a-d)a(a+d)=440...\left(2\right)\Rightarrow (8-d)\left(8\right)(8+d)=440\Rightarrow (8-d)(8+d)=55\Rightarrow 64-d^2=55\Rightarrow d^2=64-55=9\Rightarrow d=\pm 3$
Therefore when $d=3$, the numbers are $5,8,11$ and
when $d=-3,$ the numbers are $11,8$ and $5$.
Thus the three numbers are $5,8$ and $11.$