Three numbers are in A.P. If the sum of these numbers be 27 and the product 648,

Solution:

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

Their sum $=27$

$\Rightarrow a-d+a+a+d=27$

$\Rightarrow 3 a=27$

$\Rightarrow a=9 \quad \ldots(i)$

Product $=(a-d) a(a+d)=648$

$\Rightarrow a\left(a^{2}-d^{2}\right)=648$

$\Rightarrow 9\left(81-d^{2}\right)=648$

$\Rightarrow\left(81-d^{2}\right)=72$

$\Rightarrow d^{2}=9$

$\Rightarrow d=\pm 3$

When $\mathrm{a}=9, \mathrm{~d}=3$, we have :

$6,9,12$

When a $=9, d=-3$, we have :

$12,9,6$