Question

If the sum of first $m$ terms of an AP be $n$ and the sum of its first $n$ terms be $m$ then show that the sum of its first $m+n$ terms is $-\left( m+n \right)$\[\]

Answer 1

Hint: Use the given data to find a relation between sum of first $m$ terms and $m+n$ terms Use the sum of first $m$ terms formula of an AP to make to an equation between $m$ and $n$. Similarly, do the same for next $p$ terms. You will obtain another equation. Divide the equations accordingly to prove.

Complete step-by-step answer:
Arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP is a type of sequence where the difference between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and denoted $d$ where $d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. and ${{x}_{1}}$ is called first term.
 The sum of first $n$ terms of an AP is given by the formula
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2{{x}_{1}}+\left( n-1 \right)d \right]\]

Let us denote the sum of first $m$ terms as ${{S}_{m}}$, the sum of first $n$ terms as ${{S}_{n}}$, the sum of first $m+n$ terms as ${{S}_{m+n}}$, the sum of first $m+p$ terms as ${{S}_{m+p}}$.
We know that ${{S}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...{{x}_{m+n}}$. Let us define the next $n$ terms as ${{x}_{m+1}},{{x}_{m+2}},...,{{x}_{m+n}}$.
We are given in the question that
\[{{S}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+n}}\]
Let us add ${{S}_{m}}$ both side
\[\begin{align}
  & {{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}+{{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}+{{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+n}} \\
 & \Rightarrow {{S}_{m}}+{{S}_{m}}={{S}_{m+n}} \\
 & \Rightarrow 2{{S}_{m}}={{S}_{m+n}} \\
\end{align}\]
Let us use for the formula for AP,
\[\begin{align}
  & 2{{S}_{m}}={{S}_{m+n}} \\
 & \Rightarrow 2\cdot \dfrac{m}{2}\left[ 2{{x}_{1}}+\left( m-1 \right)d \right]=\dfrac{m+n}{2}\left[ 2{{x}_{1}}+\left( m+n-1 \right)d \right] \\
\end{align}\]
Let us substitute $2{{x}_{1}}+\left( m-1 \right)d=k$ where $k$ is some real number.
\[\begin{align}
  & \Rightarrow 2\cdot \dfrac{m}{2}k=\dfrac{m+n}{2}\left[ k+nd \right] \\
 & \Rightarrow 2mk=\left( m+n \right)\left( k+nd \right) \\
 & \Rightarrow \left( m-n \right)k=\left( m+n \right)nd...(1) \\
\end{align}\]
Simialrly, let us define the next $p$ terms as ${{x}_{m+1}},{{x}_{m+2}},...,{{x}_{m+p}}$.
We are given in the question that
\[{{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+p}}\]
Let us add ${{S}_{m}}$ both side
\[\begin{align}
  & {{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}+{{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}+{{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+p}} \\
 & \Rightarrow {{S}_{m}}+{{S}_{m}}={{S}_{m+p}} \\
 & \Rightarrow 2{{S}_{m}}={{S}_{m+p}} \\
\end{align}\]
Let us use for the formula for AP,
\[\begin{align}
  & 2{{S}_{m}}={{S}_{m+p}} \\
 & \Rightarrow 2\cdot \dfrac{m}{2}\left[ 2{{x}_{1}}+\left( m-1 \right)d \right]=\dfrac{m+p}{2}\left[ 2{{x}_{1}}+\left( m+p-1 \right)d \right] \\
\end{align}\]
Let us substitute again $2{{x}_{1}}+\left( m-1 \right)d=k$
\[\begin{align}
  & \Rightarrow 2\cdot \dfrac{m}{2}k=\dfrac{m+p}{2}\left[ k+pd \right] \\
 & \Rightarrow 2mk=\left( m+p \right)\left( k+pd \right) \\
 & \Rightarrow \left( m-p \right)k=\left( m+p \right)pd...(2) \\
\end{align}\]
Dividing equation (1) with equation (2) we get
\[\begin{align}
  & \dfrac{\left( m-n \right)k}{\left( m-p \right)k}=\dfrac{\left( m+n \right)kd}{\left( m+p \right)pd} \\
 & \Rightarrow \left( m-n \right)\left( m+p \right)p=\left( m-p \right)\left( m+n \right)k \\
\end{align}\]
Dividing both side by $mnp$ we get,
\[\left( \dfrac{1}{n}-\dfrac{1}{m} \right)\left( m+n \right)=\left( m+p \right)\left( \dfrac{1}{p}-\dfrac{1}{m} \right)\]
Taking negative sign both side we get,
\[\left( m+n \right)\left( \dfrac{1}{m}-\dfrac{1}{n} \right)=\left( m+p \right)\left( \dfrac{1}{m}-\dfrac{1}{p} \right)\]

Note: The key in solving this question is design of equations using the data and known formula. We also need to take care order division of equations. It is assumed in the solution that all the terms which we have divided are non-zero like $k$.