Question

The sum and difference of A.M. and G.M. of two observations are $ 27 $ and $ 3 $ respectively. Find the observations.

Answer 1

Hint: In this problem, we need to find two observations such that the sum of their arithmetic mean (A.M.) and geometric mean (G.M.) will be $ 27 $ and the difference of their A.M. and G.M. will be $ 3 $ . For this, first we will write the formula of A.M. and G.M. of two assumed numbers say $ x $ and $ y $ . Then, we will use a simple elimination method to find values of $ x $ and $ y $ .

Complete step-by-step answer:
Let us assume that $ x $ and $ y $ are two required observations. We know that arithmetic mean (A.M.) of two observations $ x $ and $ y $ is given by $ \dfrac{{x + y}}{2} $ . Also we know that the geometric mean (G.M.) of two observations $ x $ and $ y $ is given by $ \sqrt {xy} $ .
In the problem, it is given that the sum of A.M. and G.M. of two observations $ x $ and $ y $ is $ 27 $ . So, we can write $ \left( {\dfrac{{x + y}}{2}} \right) + \left( {\sqrt {xy} } \right) = 27 \cdots \cdots \left( 1 \right) $ .
Let us simplify the equation $ \left( 1 \right) $ . Therefore, we get
  $
  x + y + 2\sqrt {xy} = 2\left( {27} \right) \\
   \Rightarrow x + 2\left( {\sqrt x } \right)\left( {\sqrt y } \right) + y = 54 \cdots \cdots \left( 2 \right) \\
   $
We know that $ {\left( {\sqrt x } \right)^2} = x $ and $ {\left( {\sqrt y } \right)^2} = y $ . Use this information on LHS of the equation $ \left( 2 \right) $ . Therefore, we get $ {\left( {\sqrt x } \right)^2} + 2\left( {\sqrt x } \right)\left( {\sqrt y } \right) + {\left( {\sqrt y } \right)^2} = 54 \cdots \cdots \left( 3 \right) $ . Also we know that $ {a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2} $ . Use this information on LHS of the equation $ \left( 3 \right) $ . So, we get
  $
  {\left( {\sqrt x + \sqrt y } \right)^2} = 54 \\
   \Rightarrow \sqrt x + \sqrt y = \sqrt {54} \\
   \Rightarrow \sqrt x + \sqrt y = \sqrt {9 \times 6} \\
   \Rightarrow \sqrt x + \sqrt y = 3\sqrt 6 \cdots \cdots \left( 4 \right) \\
   $
In the problem, it is also given that the difference of A.M. and G.M. of two observations $ x $ and $ y $ is $ 3 $ . So, we can write $ \left( {\dfrac{{x + y}}{2}} \right) - \left( {\sqrt {xy} } \right) = 3 \cdots \cdots \left( 5 \right) $ .
Let us simplify the equation $ \left( 5 \right) $ . Therefore, we get
  $
  x + y - 2\sqrt {xy} = 2\left( 3 \right) \\
   \Rightarrow x - 2\left( {\sqrt x } \right)\left( {\sqrt y } \right) + y = 6 \\
   \Rightarrow {\left( {\sqrt x } \right)^2} - 2\left( {\sqrt x } \right)\left( {\sqrt y } \right) + {\left( {\sqrt y } \right)^2} = 6 \cdots \cdots \left( 6 \right) \\
   $
Also we know that $ {a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2} $ . Use this information on LHS of the equation $ \left( 6 \right) $ . So, we get
  $
  {\left( {\sqrt x - \sqrt y } \right)^2} = 6 \\
   \Rightarrow \sqrt x - \sqrt y = \sqrt 6 \cdots \cdots \left( 7 \right) \\
   $
Now we will add equations $ \left( 4 \right) $ and $ \left( 7 \right) $ to eliminate $ y $ . Therefore, we get
  $
  \sqrt x + \sqrt y + \sqrt x - \sqrt y = 3\sqrt 6 + \sqrt 6 \\
   \Rightarrow 2\sqrt x = 4\sqrt 6 \\
   \Rightarrow {\left( {2\sqrt x } \right)^2} = {\left( {4\sqrt 6 } \right)^2} \\
   \Rightarrow 4x = 16 \times 6 \\
   \Rightarrow x = \dfrac{{16 \times 6}}{4} \\
   \Rightarrow x = 24 \\
   $
Now we will substitute $ x = 24 $ in the equation $ \left( 4 \right) $ to find the value of $ y $ . Therefore, we get
  $
  \sqrt {24} + \sqrt y = 3\sqrt 6 \\
   \Rightarrow \sqrt {4 \times 6} + \sqrt y = 3\sqrt 6 \\
   \Rightarrow \sqrt y = 3\sqrt 6 - 2\sqrt 6 \\
   \Rightarrow \sqrt y = \sqrt 6 \\
   \Rightarrow {\left( {\sqrt y } \right)^2} = {\left( {\sqrt 6 } \right)^2} \\
   \Rightarrow y = 6 \\
   $
Therefore, we can say that the required observations are $ 24 $ and $ 6 $ .

Note: If there are $ n $ positive numbers say $ {a_1},{a_2}, \ldots ,{a_n} $ then arithmetic mean (A.M.) is obtained by using the formula $ \dfrac{{{a_1} + {a_2} + \ldots + {a_n}}}{n} $ and geometric mean is obtained by using the formula $ \sqrt[n]{{{a_1} \times {a_2} \times \ldots \times {a_n}}} $ . The sum of A.M. and G.M. is always positive. Also the difference between A.M. and G.M. is always positive.