Question

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

Answer 1

Step 1: Finding values of a and d

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

The sum of the numbers is 27:
$$\begin{gathered} a-d+a+a+d=27 \\ \Rightarrow 3a=27 \\ \Rightarrow a=9 \end{gathered}$$

The product of numbers is 648:
$$\begin{gathered} (a-d)\left(a\right)(a+d)=648 \\ a\left(a+d\right)\left(a-d\right)=648 \\ \Rightarrow a(a^2-d^2)=648 \\ \Rightarrow 9(9^2-d^2)=648 \\ \Rightarrow 81-d^2=72 \\ \Rightarrow d^2=81-72 \\ \Rightarrow d^2=9 \\ \Rightarrow d=\pm 3 \end{gathered}$$

Step 2: Computing numbers $a-d,a,a+d$ for $d=3$:

$$\begin{gathered} a-d=9-3=6 \\ a=9 \\ a+d=9+3=12 \end{gathered}$$

Step 3: Computin numbers $a-d,a,a+d$ for $d=-3$:

$$\begin{gathered} a-d=9+3=12 \\ a=9 \\ a+d=9-3=6 \end{gathered}$$

Hence, the numbers are $6,9,12$ or $12,9,6$.