Three numbers are in AP. If their sum is 27 and their product is 648,

To Find: The three numbers which are in AP.

Given: Sum and product of three numbers are 27 and 648 respectively.

Let required number be $(a-d),(a),(a+d)$. Then,

$(a-d)+a+(a+d)=27 \Rightarrow 3 a=27 \Rightarrow a=9$

Thus, the numbers are $(9-d), 9$ and $(9+d)$

But their product is 648 .

$\therefore(9-d) \times 9 \times(9+d)=648$

$\Rightarrow(9-d)(9+d)=72$

$\Rightarrow 81-d^{2}=72 \Rightarrow d^{2}=9 \Rightarrow d=\pm 3$

When $d=3$ numbers are $6,9,12$

When $d=(-3)$ numbers are $12,9,6$

So, Numbers are $6,9,12$ or $12,9,6$.