Question

If ${{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}}$ are the arithmetic means inserted between the numbers $a$ and $b$, then find the value of $\left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)$

Answer 1

Hint: In this problem we have an arithmetic progression ${{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}}$which is inserted between the numbers $a$ and $b$. Now the Arithmetic Progression will become $a,{{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}},b$. In the above A.P we will calculate the value of common difference$\left( d \right)$ of the A.P by using the formula for ${{n}^{th}}$ term of A.P i.e. ${{a}_{n}}=a+\left( n-1 \right)d$, where $a$ is the first term of the A.P and the value ${{a}_{n}}=b$. so, we will calculate the both the required values from the series $a,{{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}},b$ and then we will find the value of $b$. After calculating the value of $d$, we will find the values of ${{A}_{51}}$, ${{A}_{1}}$. From those values we will calculate the required value.

Complete step-by-step answer:
Given that, the series ${{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}}$ is inserted between the numbers $a$ and $b$, then the series becomes as $a,{{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}},b$.
In the above A.P the first term is $a=a$, final term/ ${{m}^{th}}$ term is ${{a}_{m}}=b$.
We know that in A.P, the value of ${{n}^{th}}$ term is given by ${{a}_{n}}=a+\left( n-1 \right)d$, then
$\begin{align}
  & {{a}_{m}}=a+\left( m-1 \right)d \\
 & \Rightarrow b=a+\left( m-1 \right)d \\
 & \Rightarrow b-a=\left( m-1 \right)d \\
 & \Rightarrow d=\dfrac{b-a}{m+1} \\
\end{align}$
Here we have the last term ${{A}_{51}}$, then $m=51$
$\begin{align}
  & \therefore d=\dfrac{b-a}{51+1} \\
 & \Rightarrow d=\dfrac{b-a}{52} \\
\end{align}$
Now the values of ${{A}_{51}}$, ${{A}_{1}}$ are
${{A}_{51}}=a+\left( 52-1 \right)d$
Substituting the value of $d$, then we will get
$\begin{align}
  & {{A}_{51}}=a+\left( 52-1 \right)d \\
 & \Rightarrow {{A}_{51}}=a+51\left( \dfrac{b-a}{52} \right) \\
 & \Rightarrow {{A}_{51}}=\dfrac{52a+51b-51a}{52} \\
 & \Rightarrow {{A}_{51}}=\dfrac{a+51b}{52} \\
\end{align}$
Similarly, the value of ${{A}_{1}}$ is given by
${{A}_{1}}=a+\left( 2-1 \right)d$
Substituting the value of $d$in the above equation then we will get
$\begin{align}
  & {{A}_{1}}=a+\left( \dfrac{b-a}{52} \right) \\
 & \Rightarrow {{A}_{1}}=\dfrac{52a+b-a}{52} \\
 & \Rightarrow {{A}_{1}}=\dfrac{b+51a}{52} \\
\end{align}$
$\therefore $The value of $\left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)$ is
$\begin{align}
  & \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{b+\dfrac{a+51b}{52}}{b-\dfrac{a+51b}{52}}-\dfrac{\dfrac{b+51a}{52}+a}{\dfrac{b+51a}{52}-a} \\
 & \Rightarrow \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{\dfrac{52b+a+51b}{52}}{\dfrac{52b-a-51b}{52}}-\dfrac{\dfrac{b+51a+52a}{52}}{\dfrac{b+51a-52a}{52}} \\
 & \Rightarrow \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{a+103b}{b-a}-\dfrac{b+103a}{b-a} \\
\end{align}$
Taking $\dfrac{1}{\left( b-a \right)}$ as common, then we will get
$\begin{align}
  & \Rightarrow \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{1}{b-a}\left( a+103b-b-103a \right) \\
 & \Rightarrow \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{102b-102a}{b-a} \\
 & \Rightarrow \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=\dfrac{102\left( b-a \right)}{\left( b-a \right)} \\
\end{align}$
Now cancelling the term $\left( b-a \right)$, then we will get
$\therefore \left( \dfrac{b+{{A}_{51}}}{b-{{A}_{51}}} \right)-\left( \dfrac{{{A}_{1}}+a}{{{A}_{1}}+a} \right)=102$

Note: While calculating the values of ${{A}_{51}}$, ${{A}_{1}}$, students may consider the series ${{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}}$ and take ${{A}_{1}}$ as the first term, but it is actually second term in the series $a,{{A}_{1}},{{A}_{2}},{{A}_{3}},...,{{A}_{31}},b$ and the term ${{A}_{51}}$ is the ${{52}^{th}}$ term.