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

My Approach: Let the numbers be $a,a+d,a+2d$ 
So, according to question $$3a+3d=24$$ $$a+d=8$$ and $$a(a+d)(a+2d)=440$$ $$8a+ad=55$$ I can’t proceed from here. 
Please help.
 A: Let the numbers be : $a-d, a, a+d$ 
Then According To Question.  $$a-d+a+a+d=24$$ $$a=8$$ and $$(a-d)(a)(a+d)=440$$ $$(8-d)(8)(8+d)=440$$ $$(8-d)(8+d)=55$$ $$64-d^2=55$$ $$-d^2=-9$$ $$d=\pm3$$ $\therefore$ the A.P. is $$5,8,11$$ $$**OR**$$ $$11,8,5$$
A: Whenever You struggle with such problems, choose such a sequence so that you can get at least the value of first term. Notice that If you will choose an AP such that three terms are $a-b,a,a+b$, on adding them, all the $b's$ will cancel out and You will be left with $a$.
So, $a-b+a+a+b=24\implies 3a=24\implies a=8$. And then You will deal with something like$$(a-b)(a)(a+b)=440\implies (8-b)(b)(8+b)$$
Now, you have a quadratic equation, solve it and get the answer.
A: Another way to view it is let $m$ equal the middle term.
So they are $m -d, m, m+d$
$(m-d) + m + (m+d) = 3m = 24$ so $m = 8$.
So the numbers are $8-d, 8, 8+d$ and
$(8-d)8(8-d) = 8(64 -d^2)=440$
$64 -d^2 = 55$
$d^2 = 64 - 55 = 9$
$d = \pm 3$
so the numbers are $5,8,11$. Or $11, 8, 5$
Even if we did it your way.
$a + d = 8$ and $8a + ad = 55$ we'd have.
$d = 8 -a$
$8a + a(8-a) = 8a +8a - a^2 = 55$
$a^2 - 16a + 55 = 0$
so
$a = \frac {16 \pm \sqrt{16^2 - 4*55}}{2} = $
$\frac {16 \pm \sqrt {256 - 220}}{2} = 8 \pm \frac{\sqrt{36}}2 = 8 \pm 3 = 5;11$
so $d = 8 -5 = 3$ or $d = 8 -11 = -3$
And the numbers are $5, 8, 11$ or $11, 8, 5$.
