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Question: If the sum of first \(m\) terms of an AP be \(n\) and the sum of its first \(n\) terms be \(m\) then...

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

Explanation

Solution

Use the given data to find a relation between sum of first mm terms and m+nm+n terms Use the sum of first mm terms formula of an AP to make to an equation between mm and nn. Similarly, do the same for next pp 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 (xn)=x1,x2,x3,...\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},... is an AP, then x2−x1=x3−x2...{{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}... . The difference between two terms is called common difference and denoted dd where d=x2−x1=x3−x2...d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}.... and x1{{x}_{1}} is called first term.
The sum of first nn terms of an AP is given by the formula
Sn=n2[2x1+(n−1)d]{{S}_{n}}=\dfrac{n}{2}\left[ 2{{x}_{1}}+\left( n-1 \right)d \right]

Let us denote the sum of first mm terms as Sm{{S}_{m}}, the sum of first nn terms as Sn{{S}_{n}}, the sum of first m+nm+n terms as Sm+n{{S}_{m+n}}, the sum of first m+pm+p terms as Sm+p{{S}_{m+p}}.
We know that Sm=x1+x2+....+xm=xm+1+xm+2+...xm+n{{S}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...{{x}_{m+n}}. Let us define the next nn terms as xm+1,xm+2,...,xm+n{{x}_{m+1}},{{x}_{m+2}},...,{{x}_{m+n}}.
We are given in the question that
Sm=x1+x2+....+xm=xm+1+xm+2+...+xm+n{{S}_{m}}={{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+n}}
Let us add Sm{{S}_{m}} both side

& {{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{aligned}$$ Let us use for the formula for AP, $$\begin{aligned} & 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{aligned}$$ Let us substitute $2{{x}_{1}}+\left( m-1 \right)d=k$ where $k$ is some real number. $$\begin{aligned} & \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{aligned}$$ Simialrly, let us define the next $p$ terms as ${{x}_{m+1}},{{x}_{m+2}},...,{{x}_{m+p}}$. We are given in the question that $${{x}_{1}}+{{x}_{2}}+....+{{x}_{m}}={{x}_{m+1}}+{{x}_{m+2}}+...+{{x}_{m+p}}$$ Let us add ${{S}_{m}}$ both side $$\begin{aligned} & {{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{aligned}$$ Let us use for the formula for AP, $$\begin{aligned} & 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{aligned}$$ Let us substitute again $2{{x}_{1}}+\left( m-1 \right)d=k$ $$\begin{aligned} & \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{aligned}$$ Dividing equation (1) with equation (2) we get $$\begin{aligned} & \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{aligned}$$ Dividing both side by $mnp$ we get, $$\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 $k$.