Question

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P. and \[{{a}_{1}}=0\], then the value of \[\left( \dfrac{{{a}_{3}}}{{{a}_{2}}}+\dfrac{{{a}_{4}}}{{{a}_{3}}}+...+\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \right)-{{a}_{2}}\left( \dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{3}}}+...+\dfrac{1}{{{a}_{n-2}}} \right)\] is equal to
A. $\left( n-2 \right)+\dfrac{1}{\left( n-2 \right)}$
B. $\dfrac{1}{\left( n-2 \right)}$
C. $\left( n-2 \right)$
D. $n+2$

Answer 1

Hint: We express the arithmetic sequence in its general form and find the terms with respect to the common difference. We put the values and simplify them to find the sum of 1s for $\left( n-3 \right)$ times. We try to form the expression with the given options.

Complete step-by-step solution:
We express the arithmetic sequence \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] in its general form.
We express the terms as ${{a}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term is \[{{a}_{1}}=0\] and the common difference be $d$ where $d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}$.
We can express the general term ${{t}_{n}}$ based on the first term and the common difference.
The formula being ${{a}_{n}}={{a}_{n-1}}+d$.
Therefore,
$\begin{align}
  & {{a}_{2}}=d+{{a}_{1}}=d \\
 & {{a}_{3}}=d+{{a}_{2}}=d+d=2d \\
\end{align}$
And so on.
The general form becomes \[{{a}_{n}}=\left( n-1 \right)d\].
We now put the values in the equation \[\left( \dfrac{{{a}_{3}}}{{{a}_{2}}}+\dfrac{{{a}_{4}}}{{{a}_{3}}}+...+\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \right)-{{a}_{2}}\left( \dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{3}}}+...+\dfrac{1}{{{a}_{n-2}}} \right)\].
\[\begin{align}
  & \left( \dfrac{{{a}_{3}}}{{{a}_{2}}}+\dfrac{{{a}_{4}}}{{{a}_{3}}}+...+\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \right)-{{a}_{2}}\left( \dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{3}}}+...+\dfrac{1}{{{a}_{n-2}}} \right) \\
 & =\left( \dfrac{2d}{d}+\dfrac{3d}{2d}+...+\dfrac{\left( n-1 \right)d}{\left( n-2 \right)d} \right)-d\left( \dfrac{1}{d}+\dfrac{1}{2d}+...+\dfrac{1}{\left( n-3 \right)d} \right) \\
\end{align}\]
We now omit the value of $d$ to get
\[\begin{align}
  & \left( \dfrac{2d}{d}+\dfrac{3d}{2d}+...+\dfrac{\left( n-1 \right)d}{\left( n-2 \right)d} \right)-d\left( \dfrac{1}{d}+\dfrac{1}{2d}+...+\dfrac{1}{\left( n-3 \right)d} \right) \\
 & =\left( 2+\dfrac{3}{2}+\dfrac{4}{3}+...+\dfrac{\left( n-2 \right)}{\left( n-3 \right)}+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \right)-\left( 1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{\left( n-3 \right)} \right) \\
\end{align}\]
We now take the respective terms of the two summations in a particular form
\[\begin{align}
& \left( 2+\dfrac{3}{2}+\dfrac{4}{3}+...+\dfrac{\left( n-2 \right)}{\left( n-3 \right)}+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \right)-\left( 1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{\left( n-3 \right)} \right) \\
 & =\left( 2-1 \right)+\left( \dfrac{3}{2}-\dfrac{1}{2} \right)+\left( \dfrac{4}{3}-\dfrac{1}{3} \right)+....+\left( \dfrac{\left( n-2 \right)}{\left( n-3 \right)}-\dfrac{1}{\left( n-3 \right)} \right)+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \\
\end{align}\]
We get 1 in every subtraction. We add them.
\[\begin{align}
  & \left( 2-1 \right)+\left( \dfrac{3}{2}-\dfrac{1}{2} \right)+\left( \dfrac{4}{3}-\dfrac{1}{3} \right)+....+\left( \dfrac{\left( n-2 \right)}{\left( n-3 \right)}-\dfrac{1}{\left( n-3 \right)} \right)+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \\
 & =\underbrace{1+1+1+...+1}_{\left( n-3 \right)times}+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \\
 & =\left( n-3 \right)+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \\
\end{align}\]
We now subtract 1 from the former one and try to equate with the options.
\[\begin{align}
  & \left( n-3 \right)+\dfrac{\left( n-1 \right)}{\left( n-2 \right)} \\
 & =\left( n-2 \right)+\left[ \dfrac{\left( n-1 \right)}{\left( n-2 \right)}-1 \right] \\
 & =\left( n-2 \right)+\dfrac{\left( n-1 \right)-n+2}{\left( n-2 \right)} \\
 & =\left( n-2 \right)+\dfrac{1}{\left( n-2 \right)} \\
\end{align}\]
Therefore, \[\left( \dfrac{{{a}_{3}}}{{{a}_{2}}}+\dfrac{{{a}_{4}}}{{{a}_{3}}}+...+\dfrac{{{a}_{n}}}{{{a}_{n-1}}} \right)-{{a}_{2}}\left( \dfrac{1}{{{a}_{2}}}+\dfrac{1}{{{a}_{3}}}+...+\dfrac{1}{{{a}_{n-2}}} \right)\] is equal to $\left( n-2 \right)+\dfrac{1}{\left( n-2 \right)}$.
The correct option is A.

Note: The sequence is an increasing sequence where the common difference is a positive number. The common difference will never be calculated according to the difference of greater number from the lesser number.