Question

Find the number of triplets of integers in arithmetic progression, the sum of whose squares is 1994
a. 36
b. 45
c. 12
d. Does not exist

Answer 1

Hint: To solve the given question we need to know how the numbers are arranged in an arithmetic progression (AP). The numbers in an AP, will have a common difference that is, if a is the first number in an AP, then the arithmetic progression will be, a, a+r, a+2r, a+3r, … .

Complete step-by-step answer:
In this question we have been asked to find the number of triplets of integers, triplet refers to a set of 3 numbers. Also, we have been given that the numbers are taken from an arithmetic progression. Therefore, let us take the numbers as a-r, a and a+r. Here r is the common difference and a and r are both integers.
Another condition given in the question is that the sum of squares of the triplets should be equal to 1994. So, we can write this condition as follows.
 $ {{\left( a-r \right)}^{2}}+{{a}^{2}}+{{\left( a+r \right)}^{2}}=1994 $
Now, expanding the brackets using the identities $ {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}and{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ we get,
 $ \left( {{a}^{2}}-2ar+{{r}^{2}} \right)+\left( {{a}^{2}} \right)+\left( {{a}^{2}}+2ar+{{r}^{2}} \right)=1994 $
On solving it further, we will get,
 $ 3{{a}^{2}}+2{{r}^{2}}=1994 $
On transposing $ 3{{a}^{2}} $ from LHS to the RHS, we get,
 $ 2{{r}^{2}}=1994-3{{a}^{2}} $
On dividing the whole equation by 2, we get,
 $ {{r}^{2}}=\dfrac{1994-3{{a}^{2}}}{2} $
We can also write it as,
 $ \begin{align}
  & {{r}^{2}}=\dfrac{1994}{2}-\dfrac{3{{a}^{2}}}{2} \\
 & {{r}^{2}}=997-\dfrac{3{{a}^{2}}}{2} \\
\end{align} $
But in the question it is given that both a and r are integers, that is,
 $ r\in I\Rightarrow {{r}^{2}}\in I $ .
Now, on substituting the value of $ {{r}^{2}} $ as $ 997-\dfrac{3{{a}^{2}}}{2} $ in the above expression, we will get,
 $ 997-\dfrac{3{{a}^{2}}}{2}\in I $
This can happen only if $ {{a}^{2}} $ is an integer that cancels out the fraction $ \dfrac{3}{2} $ , which means that $ {{a}^{2}} $ should be an integer of the form 2n.
So, \[{{a}^{2}}\in Even\text{ }number\], which implies that \[a\in Even\text{ }number\]. If a is even, then let us suppose, a = 2n for some n = 0, 1, 2, …. So, we can write, $ {{a}^{2}}=4{{n}^{2}} $
Now, the value of $ {{r}^{2}} $ can be re-written as,
 $ \begin{align}
  & {{r}^{2}}=997-\dfrac{3}{2}\left( 4{{n}^{2}} \right) \\
 & {{r}^{2}}=997-3\left( 2{{n}^{2}} \right) \\
 & {{r}^{2}}=997-6{{n}^{2}}.........(i) \\
\end{align} $
But we know that the square of a number is always positive, that is $ {{r}^{2}}\ge 0 $ . So, we can write equation (i) as,
 $ 997-6{{n}^{2}}\ge 0 $
On adding $ 6{{n}^{2}} $ on both the sides, we get,
 $ \begin{align}
  & 997\ge 6{{n}^{2}} \\
 & 6{{n}^{2}}\le 997 \\
\end{align} $
On dividing the above expression by 6, we get,
 $ \begin{align}
  & {{n}^{2}}\le \dfrac{997}{6} \\
 & {{n}^{2}}\le 166.16 \\
\end{align} $
On taking the square root on both the sides, we get,
 $ n\le 12.89 $
So, now we know that n can take values from 0 up to 12.89.
Therefore, the values of n = 0, 1, 2, 3, …. 12.
Now, let us observe the values of $ {{r}^{2}} $ corresponding to the value of n. So, on substituting the values of n in equation (i), we get,
For n = 0, we get $ {{r}^{2}}=997-6\times 0=997 $ . And we know that 997 is not a perfect square. So, $ r\notin I $ .
For n = 1, we get, $ {{r}^{2}}=997-6\times {{1}^{2}}=997-6=996 $ . 996 is also not a perfect square. So, $ r\notin I $ .
For n = 2, we get, $ {{r}^{2}}=997-6\times {{2}^{2}}=997-24=973 $ . 973 is also not a perfect square. So, $ r\notin I $ .
For n = 3, we get, $ {{r}^{2}}=997-6\times {{3}^{2}}=997-54=943 $ . 943 is also not a perfect square. So, $ r\notin I $ .
On continuing these substitutions till n = 12, we will not be able to get the value of r as an integer.
Therefore, the number of triplets of integers in arithmetic progression, the sum of whose squares is 1994 is 0.
Hence, option (d) is the correct answer, that is, no such triplets exist.

Note: Here, we have taken the numbers in the AP as a-r, a and a+r to cancel out the term 2ar while substituting n in the equation. If such a term exists, the calculation would have been different and it will be more complicated, hence, choose the variables suitably for your solution.