Question

The relation between x and y in order that the 20th arithmetic mean between x and 2y is same as the 20th arithmetic mean between 2x and y (99 means being inserted in each case) is:

Answer 1

Hint: We will use the concept of Arithmetic Progression to solve this question. We will use the formula \[{A_r} = a + rd\] to find the rth arithmetic mean where \[{A_r}\] is the arithmetic mean, a is the first term, r is the number of the term to be found and d is the common difference between two terms throughout the series.

Complete step-by-step answer:
Firstly, we will let \[x,{A_1},{A_2},{A_3},....,{A_{99}},2y\] be an AP series.
The common difference between x and 2y will be \[d = \dfrac{{2y - x}}{{n + 1}}\]where n is the total number of means inserted between the two numbers.
We will now find the 20th arithmetic mean using the formula \[{A_r} = a + rd\]
\[
  {A_{20}} = a + 20d \\
   \Rightarrow {A_{20}} = x + 20\left( {\dfrac{{2y - x}}{{99 + 1}}} \right) \\
   \Rightarrow {A_{20}} = x + 20\left( {\dfrac{{2y - x}}{{100}}} \right) \\
   \Rightarrow {A_{20}} = x + \left( {\dfrac{{2y - x}}{5}} \right) \\
   \Rightarrow {A_{20}} = \dfrac{{5x + 2y - x}}{5} \\
   \Rightarrow {A_{20}} = \dfrac{{4x + 2y}}{5} \\
    \\
\]
Now, we will let \[2x,{A_1},{A_2},{A_3},....,{A_{99}},y\] be an AP series.
The common difference between 2x and y will be \[d = \dfrac{{y - 2x}}{{n + 1}}\]where n is the total number of means inserted between the two numbers.
We will now find the 20th arithmetic mean using the formula \[{A_r} = a + rd\]
\[
  {A_{20}} = a + 20d \\
   \Rightarrow {A_{20}} = 2x + 20\left( {\dfrac{{y - 2x}}{{99 + 1}}} \right) \\
   \Rightarrow {A_{20}} = 2x + 20\left( {\dfrac{{y - 2x}}{{100}}} \right) \\
   \Rightarrow {A_{20}} = 2x + \left( {\dfrac{{y - 2x}}{5}} \right) \\
   \Rightarrow {A_{20}} = \dfrac{{10x + y - 2x}}{5} \\
   \Rightarrow {A_{20}} = \dfrac{{8x + y}}{5} \\
\]
According to the question, both the 20th arithmetic means are equal.
\[
  \therefore \dfrac{{4x + 2y}}{5} = \dfrac{{8x + y}}{5} \\
   \Rightarrow 4x + 2y = 8x + y \\
   \Rightarrow 2y - y = 8x - 4x \\
   \Rightarrow y = 4x \\
   \Rightarrow x = \dfrac{y}{4} \\
\]
Therefore, the relation between x and y in order that the 20th arithmetic mean between x and 2y is the same as the 20th arithmetic mean between 2x and y is \[x = \dfrac{y}{4}\].

Note: We are calculating the common difference by subtracting the first term from the last term and dividing it by the sum of the total terms inserted and 1. We are not using the formula for the nth term of the AP instead we are using \[{A_r} = a + rd\].