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

Solution:

Let the three numbers be $(a-d), a,(a+d)$.

Sum $=24$

$\Rightarrow(a-d)+a+(a+d)=24$

$\Rightarrow 3 a=24$

$\Rightarrow a=8 \quad \ldots(i)$

Product $=a(a-d)(a+d)=440$

$\Rightarrow a\left(a^{2}-d^{2}\right)=440$

$\Rightarrow 8\left(64-d^{2}\right)=440$    (From (i))

$\Rightarrow\left(64-d^{2}\right)=55$

$\Rightarrow d^{2}=9$

$\Rightarrow d=\pm 3$

With $\mathrm{a}=8, \mathrm{~d}=3$, we have :

$5,8,11$

With a $=8, \mathrm{~d}=-3$, we have:

$11,8,5$