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**Hint:**To solve this question we will first calculate the sum of first m terms of the A.P. using the formula $S=\dfrac{m}{2}\left( 2a+\left( m-1 \right)d \right)$ where m is the number of terms and a is the first term of the A.P. and d is the common difference. Similarly we will find the sum of first n terms of the A.P. and then equate the both obtained sum as given in the question. From there we will get some conditions that we will use to find the sum of first (m+n) terms and will prove it to be zero using those conditions.

**Complete step by step answer:**

To solve this question we need to know that we can calculate the sum of the first n terms of an A.P. using the formula $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where a is the first term of the A.P. and d is the common difference of the A.P.,

So the sum of the first m terms of the A.P. we will get as,

= $\dfrac{m}{2}\left( 2a+\left( m-1 \right)d \right)$

And similarly we will find the the sum of the first n terms using this formula,

$=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$

Now according to the question sum of the first m terms is equal to the sum of the first n terms hence equating both the obtained sum we get,

$\dfrac{m}{2}\left( 2a+\left( m-1 \right)d \right)=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$

Cancelling 2 from denominator on both sides, we get

$\begin{align}

& \Rightarrow m\left( 2a+\left( m-1 \right)d \right)=n\left( 2a+\left( n-1 \right)d \right) \\

& \Rightarrow 2am+md\left( m-1 \right)=2an+nd\left( n-1 \right) \\

& \Rightarrow 2am-2an=nd\left( n-1 \right)-md\left( m-1 \right) \\

\end{align}$

Taking d common in the RHS of the equation we get,

$\begin{align}

& \Rightarrow 2a\left( m-n \right)=d\left[ n\left( n-1 \right)-m\left( m-1 \right) \right] \\

& \Rightarrow 2a\left( m-n \right)=d\left[ {{n}^{2}}-n-\left( {{m}^{2}}-m \right) \right] \\

& \Rightarrow 2a\left( m-n \right)=d\left[ {{n}^{2}}-n-{{m}^{2}}+m \right] \\

\end{align}$

$\Rightarrow 2a\left( m-n \right)=-d\left[ \left( {{m}^{2}}-{{n}^{2}} \right)-\left( m-n \right) \right]$

We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, so we get

$\begin{align}

& \Rightarrow 2a\left( m-n \right)=-d\left[ \left( m-n \right)\left( m+n \right)-\left( m-n \right) \right] \\

& \Rightarrow 2a\left( m-n \right)=-d\left( m-n \right)\left[ m+n-1 \right] \\

\end{align}$

Now taking the RHS to LHS we get,

$\begin{align}

& \Rightarrow 2a\left( m-n \right)+d\left( m-n \right)\left[ m+n-1 \right]=0 \\

& \Rightarrow \left( m-n \right)\left[ 2a+d\left( m+n-1 \right) \right]=0 \\

\end{align}$

So,

Either we have,

$\begin{align}

& m-n=0 \\

& m=n \\

\end{align}$

But we know that m can’t be equal to n as both are different number of terms,

So we get,

$2a+d\left( m+n-1 \right)=0\,\,\,......\left( 1 \right)$

Now we will find the sum of the first (m+n) terms as,

${{S}_{m+n}}=\dfrac{m+n}{2}\left( 2a+\left( \left( m+n \right)-1 \right)d \right)$

Now using equation 1 we have $2a+d\left( m+n-1 \right)=0\,$, so we get

$\begin{align}

& {{S}_{m+n}}=\dfrac{m+n}{2}\times 0 \\

& {{S}_{m+n}}=0 \\

\end{align}$

Hence we proved the sum of m+n terms to be zero.

**Note:**To solve this problem students should have prior knowledge of how to calculate the sum of the first n terms of an A.P. and in these type of problems you have to solve the mentioned data in the question to get some useful condition that we can directly use to solve the problem. Also remember the formula mentioned for the sum of the first n terms of A.P. for more problems like this.

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