Question

The sum of three numbers in Ap is $12$ and the sum of their cubes is $288$. Find the numbers.$3a^3+6a{d}^2=288$

Answer 1

Step 1. Take three consecutive terms of an A.P.

Given that there are three numbers which are in A.P. and the sum of the numbers is $12$.

The sum of the cube of the numbers is $288$.

Three consecutive terms in an A.P. are $a-d,a,a+d$.

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

Step 2. Find the sum of the numbers

The sum of the numbers are $a-d+a+a+d$

$=3a$

According to the question

$$\begin{gathered} 3a=12 \\ \Rightarrow a=4 \end{gathered}$$

Step 3. Find the sum of the cube of the numbers

The sum of the cube of the numbers is $={\left(a-d\right)}^3+a^3+{\left(a+d\right)}^3$

$$\begin{gathered} =a^3-3a^{2}d+3a{d}^2-d^3+a^3+a^3+3a^{2}d+3a{d}^2+d^3 \\ =3a^3+6a{d}^2 \end{gathered}$$

According to the question,

$3a^3+6a{d}^2=288$

Putting the value of $a$

$3·4^3+6·4·d^2=288$

$$\begin{gathered} \Rightarrow 192+24d^2=288 \\ \Rightarrow 24d^2=288-192 \\ \Rightarrow 24d^2=96 \\ \Rightarrow d^2=4 \\ \Rightarrow d=\pm 2 \end{gathered}$$

Step 4. Finding the three numbers

For $a=4$ and $d=2$, the three numbers of the A.P. are

$$\begin{gathered} a-d=4-2 \\ =2 \end{gathered}$$

$a=4$

$$\begin{gathered} a+d=4+2 \\ =6 \end{gathered}$$

For $a=4$ and $d=-2$, the three numbers of the A.P. are

$$\begin{gathered} a-d=4+2 \\ =6 \end{gathered}$$

$a=4$

$$\begin{gathered} a+d=4-2 \\ =2 \end{gathered}$$

Hence, the three numbers are $2,4,6$.