Question

The sum of three numbers in A.P is 12, and the sum of their cube is 288. Find the number.

Answer 1

Hint: In this question, we have to find three numbers whose sum and sum of their cubes is given. We are also given that they are in A.P, so we will suppose one of them as variable and take common difference (d) for other two and suppose there numbers in two variables. Using sum and sum of their cubes condition, we will form two equations and thus solve them to find three numbers. We will use following two operations:
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\text{ and }{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3a{{b}^{2}}+3{{a}^{2}}b\]

Complete step-by-step solution:
Here, we are given three numbers who are in A.P. Let us suppose one of the numbers as 'a'. Let us suppose there is a common difference between any two numbers in an A.P as 'd'. Then, a+d will be the serial number. Since, the difference can be subtracted too, so that, we can get the third number. So, a-d will be the third number.
Thus, we have obtained three numbers a-d, a, a+d which are in A.P.
Now, we are given sum of these numbers as 12, so let us add all three numbers, we get:
\[\begin{align}
  & a-d+a+a+d=12 \\
 & \Rightarrow 3a=12 \\
 & \Rightarrow a=4\cdots \cdots \cdots \cdots \cdots \left( 1 \right) \\
\end{align}\]
Hence, we get one of the number as 4. But to find the other two, we need to find the value of d.
Now, we are given sum of the cube of these three numbers, so let us use this, so we get:
\[{{\left( a-d \right)}^{3}}+{{a}^{3}}+{{\left( a+d \right)}^{3}}=288\]
As we know, \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\text{ and }{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3a{{b}^{2}}+3{{a}^{2}}b\]. So, using this formula in above equation we get:
\[{{a}^{3}}-{{d}^{3}}-3{{a}^{2}}d+3a{{d}^{2}}+{{a}^{3}}+{{a}^{3}}+{{d}^{3}}+3{{a}^{2}}d+3a{{d}^{2}}=288\]
As we can see, ${{d}^{3}},3{{a}^{2}}d$ cancel out, so we get:
\[\begin{align}
  & {{a}^{3}}+3a{{d}^{2}}+{{a}^{3}}+{{a}^{3}}+3a{{d}^{2}}=288 \\
 & \Rightarrow 3{{a}^{3}}+6a{{d}^{2}}=288 \\
\end{align}\]
As we have calculated earlier the value of 'a', we can put it in above equation to get value of d, so putting a=4 from (1) in above equation, we get:
\[\begin{align}
  & 3{{\left( 4 \right)}^{3}}+6\left( 4 \right){{d}^{2}}=288 \\
 & \Rightarrow 192+24{{d}^{2}}=288 \\
 & \Rightarrow 24{{d}^{2}}=196 \\
 & \Rightarrow {{d}^{2}}=4 \\
 & \Rightarrow {{d}^{2}}=\pm 2 \\
\end{align}\]
Hence, d can be equal to 2 or -2.
Putting them both one by one, we get:
For d = 2 and a = 4, we get numbers as:
\[\begin{align}
  & 4-2,4,4+2 \\
 & \Rightarrow 2,4,6 \\
\end{align}\]
For d = -2 and a = 4, we get numbers as:
\[\begin{align}
  & 4-\left( -2 \right),4,4-2 \\
 & \Rightarrow 6,4,2 \\
\end{align}\]
Therefore, numbers are the same from both values of d. Hence, required numbers are 2, 4 and 6.

Note: Students should note that we have chosen a-d rather than a+2d, to simplify our calculations. They can also check if a-d, a, a+d forms A.P by finding the common difference that is $a-\left( a-d \right)=d\text{ and }a+d-a=d$. Hence three numbers are in A.P. Students should always consider both the value of d, as the common difference can be negative too.