Question

# Four different integers form an increasing A.P. One of these numbers is the sum of the squares of the other three numbers. Find the numbers.

Hint: We will first of all assume all the numbers as consecutive numbers of an AP and then put the values or conditions given in the question to obtain an equation then try to solve it to get the desired result.

Complete Step-by-Step solution:
Given that four different integers form an increasing Arithmetic progression.
We will assume four different integers as: a - 3d, a - d, a + d and a + 3d.
These above numbers are in the increasing order of the AP
Therefore, a + 3d is the biggest number.

Now according to the given conditions in the question one of these numbers is the sum of squares of the other three numbers.
As a + 3d is the biggest so applying the above we get,
$\begin{array}{*{35}{l}} \Rightarrow (a+3d)=\left( a-3d \right){}^\text{2}+\left( a-d \right){}^\text{2}+\left( a+d \right){}^\text{2} \\ ~ \\ \Rightarrow (a+3d)=a{}^\text{2}+9d{}^\text{2}-6ad+a{}^\text{2}+d{}^\text{2}-2ad+a{}^\text{2}+d{}^\text{2}+2ad \\ ~ \\ \Rightarrow (a+3d)=3a{}^\text{2}+11d{}^\text{2}-6ad......(i) \\ ~ \\ \Rightarrow (a+3d)=3a{}^\text{2}+11d{}^\text{2}-6ad-a-3d=0 \\ ~ \\ \Rightarrow (a+3d)=3a{}^\text{2}-a\left( 6d+1 \right)+11d{}^\text{2}-3d=0 \\ \end{array}$
For a is real so, D = b2-4ac
$\Rightarrow D=\left( 6d\text{ }+\text{ }1 \right){}^\text{2}\text{ }-\text{ }12\left( 11d{}^\text{2}\text{ }-\text{ }3d \right)\text{ }\ge \text{ }0$
$\Rightarrow D=\left( 6d\text{ }+\text{ }1 \right){}^\text{2}\text{ }-\text{ }12\left( 11d{}^\text{2}\text{ }-\text{ }3d \right)\text{ }\ge \text{ }0$
$\Rightarrow D=36d{}^\text{2}\text{ }+\text{ }12d\text{ }+\text{ }1\text{ }-\text{ }132d{}^\text{2}\text{ }+\text{ }36d\text{ }\ge \text{ }0$
$\Rightarrow D=-96d{}^\text{2}\text{ }+\text{ }48d\text{ }+\text{ }1\text{ }\ge \text{ }0$
$\Rightarrow D=96d{}^\text{2}\text{ }-\text{ }48d\text{ }-\text{ }1\text{ }\le \text{ }0$
We can choose this d from the hit and trial method.
If we choose $d=\dfrac{1}{2}$ gives
$96\left( \dfrac{1}{4} \right)-48\left( \dfrac{1}{2} \right)-1\text{ }\le \text{ }0$satisfy
Hence, the above is satisfied.
Therefore, $d=\dfrac{1}{2}$
Putting the value of d in equation (i), we get the value of a as,
$a=\dfrac{1}{2}$.
Now we will proceed to calculate the values of the numbers.
$\begin{array}{*{35}{l}} a-3d=\dfrac{1}{2}-\dfrac{3}{2}=-1 \\ ~ \\ a-d=\dfrac{1}{2}-\dfrac{1}{2}=0 \\ ~ \\ a+d=\dfrac{1}{2}+\dfrac{1}{2}=1 \\ ~ \\ a\text{ }+3d=\dfrac{1}{2}+\dfrac{3}{2}=2 \\ \end{array}$
Therefore, all the numbers are obtained as -1, 0, 1, 2.

Note: The possibility of error in this question is that the terms of the AP can be assumed without the common difference D. If there is no common difference in the terms of the AP then the obtained equation will get the wrong result because AP always works with constant common difference.