Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{4001}}$ are terms of an A.P. such that $\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+....+\dfrac{1}{{{a}_{4000}}{{a}_{4001}}}=10$ and ${{a}_{2}}+{{a}_{4000}}=50$, then find the value of $\left| {{a}_{1}}-{{a}_{4001}} \right|$.
A. 20
B. 30
C. 40
D. 50

Answer 1

Hint: From the given series of A.P. we find the general term of the series. We find the formula for ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. We put the values of ${{a}_{2}},{{a}_{4000}},{{a}_{4001}}$ in the equation of ${{a}_{2}}+{{a}_{4000}}=50$. Then we find the sum and multiplication value of ${{a}_{1}},{{a}_{4001}}$. We use the identity of \[xy={{\left( \dfrac{x+y}{2} \right)}^{2}}-{{\left( \dfrac{x-y}{2} \right)}^{2}}\] to find the solution of the problem.

Complete step-by-step solution:
We have been given a series of A.P. which is ${{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{4001}}$. There are 4001 terms in the series. We express the A.P. in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term is ${{t}_{1}}={{a}_{1}}$. Let the common difference be d.
So, $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{4001}}-{{a}_{4000}}$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}={{t}_{1}}+\left( n-1 \right)d={{a}_{1}}+d\left( n-1 \right)$.
We try to find the terms ${{a}_{2}},{{a}_{4000}},{{a}_{4001}}$.
\[{{a}_{2}}={{a}_{1}}+\left( 2-1 \right)d={{a}_{1}}+d\]
\[{{a}_{4000}}={{a}_{1}}+\left( 4000-1 \right)d={{a}_{1}}+3999d\]
\[{{a}_{4001}}={{a}_{1}}+\left( 4001-1 \right)d={{a}_{1}}+4000d\]
It’s also given ${{a}_{2}}+{{a}_{4000}}=50$ which gives
\[\begin{align}
  & {{a}_{2}}+{{a}_{4000}}=50 \\
 & \Rightarrow \left( {{a}_{1}}+d \right)+\left( {{a}_{1}}+3999d \right)=50 \\
 & \Rightarrow {{a}_{1}}+{{a}_{1}}+4000d=50 \\
 & \Rightarrow {{a}_{1}}+{{a}_{4001}}=50....(i) \\
\end{align}\]
Now we multiply d with the equation $\dfrac{1}{{{a}_{1}}{{a}_{2}}}+\dfrac{1}{{{a}_{2}}{{a}_{3}}}+....+\dfrac{1}{{{a}_{4000}}{{a}_{4001}}}=10$.
$\dfrac{d}{{{a}_{1}}{{a}_{2}}}+\dfrac{d}{{{a}_{2}}{{a}_{3}}}+....+\dfrac{d}{{{a}_{4000}}{{a}_{4001}}}=10d$.
We replace the values $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{4001}}-{{a}_{4000}}$.
$\dfrac{{{a}_{2}}-{{a}_{1}}}{{{a}_{1}}{{a}_{2}}}+\dfrac{{{a}_{3}}-{{a}_{2}}}{{{a}_{2}}{{a}_{3}}}+....+\dfrac{{{a}_{4001}}-{{a}_{4000}}}{{{a}_{4000}}{{a}_{4001}}}=10d$. Now we break the equations and get
\[\begin{align}
  & \left( \dfrac{{{a}_{2}}-{{a}_{1}}}{{{a}_{1}}{{a}_{2}}} \right)+\left( \dfrac{{{a}_{3}}-{{a}_{2}}}{{{a}_{2}}{{a}_{3}}} \right)+....+\left( \dfrac{{{a}_{4001}}-{{a}_{4000}}}{{{a}_{4000}}{{a}_{4001}}} \right)=10d \\
 & \Rightarrow \left( \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{2}}} \right)+\left( \dfrac{1}{{{a}_{2}}}-\dfrac{1}{{{a}_{3}}} \right)+.....+\left( \dfrac{1}{{{a}_{4000}}}-\dfrac{1}{{{a}_{4001}}} \right)=10d \\
\end{align}\]
The first part of a term gets cancelled out with the last part of the previous term.
\[\begin{align}
  & \Rightarrow \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{2}}}-\dfrac{1}{{{a}_{3}}}+.....+\dfrac{1}{{{a}_{4000}}}-\dfrac{1}{{{a}_{4001}}}=10d \\
 & \Rightarrow \dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{4001}}}=10d \\
\end{align}\]
Placing the values of ${{a}_{2}},{{a}_{4000}},{{a}_{4001}}$ we get
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{4001}}-{{a}_{1}}}{{{a}_{1}}.{{a}_{4001}}}=10d \\
 & \Rightarrow \dfrac{{{a}_{1}}+4000d-{{a}_{1}}}{{{a}_{1}}.{{a}_{4001}}}=10d \\
 & \Rightarrow {{a}_{1}}.{{a}_{4001}}=\dfrac{4000d}{10d}=400...(ii) \\
\end{align}\]
From equation (ii) we get
\[\begin{align}
  & {{a}_{1}}.{{a}_{4001}}=400 \\
 & \Rightarrow {{\left( \dfrac{{{a}_{1}}+{{a}_{4001}}}{2} \right)}^{2}}-{{\left( \dfrac{{{a}_{1}}-{{a}_{4001}}}{2} \right)}^{2}}=400 \\
\end{align}\]
Using the equational value of (i)
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{50}{2} \right)}^{2}}-{{\left( \dfrac{{{a}_{1}}-{{a}_{4001}}}{2} \right)}^{2}}=400 \\
 & \Rightarrow {{\left( \dfrac{{{a}_{1}}-{{a}_{4001}}}{2} \right)}^{2}}={{25}^{2}}-400=225 \\
 & \Rightarrow \left| {{a}_{1}}-{{a}_{4001}} \right|=2\sqrt{225}=30 \\
\end{align}\]
The correct option is B.

Note: We also can solve the problem using the trial and error method for the two equations. We also can assume the values of the terms ${{a}_{1}},{{a}_{4001}}$ as m and n to make the equation more understandable.