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Hint: Let us assume a variable $a$ which represents the first term of the arithmetic progression, a variable $d$ which represents the common difference of the arithmetic progression. The sum of r terms of this arithmetic progression is given by the formula ${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$. Using this formula, we can solve this question.

Before proceeding with the question, we must know the formula that will be required to solve this question.

In arithmetic progression, if we have an arithmetic progression with its first term equal to a and its common difference equal to d, then, the sum of its first r terms is given by the formula,

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

In this question, it is given that the sum of m terms of an A.P. is the same as the sum of its n terms and we have to show that the sum of its (m + n) terms is zero.

Let us assume an A.P. with its first term as a and common difference as d. Using formula (1), the sum of m terms of this A.P. is equal to,

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

Using formula (1), the sum of n terms of this A.P. is equal to,

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ . . . . . . . . . . . . (3)

Since in the question, it is given ${{S}_{m}}={{S}_{n}}$, from equation (2) and (3), we get,

\[\begin{align}

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

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

& \Rightarrow 2am+{{m}^{2}}d-md=2an+{{n}^{2}}d-nd \\

& \Rightarrow 2am-2an+{{m}^{2}}d-{{n}^{2}}d-md+nd=0 \\

& \Rightarrow 2a\left( m-n \right)+\left( {{m}^{2}}-{{n}^{2}} \right)d-d\left( m-n \right)=0 \\

\end{align}\]

There is a formula ${{m}^{2}}-{{n}^{2}}=\left( m-n \right)\left( m+n \right)$. Substituting this formula in the above equation, we get,

\[\begin{align}

& 2a\left( m-n \right)+\left( m-n \right)\left( m+n \right)d-d\left( m-n \right)=0 \\

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

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

& \Rightarrow \left( 2a+\left( m+n-1 \right)d \right)=0..................\left( 4 \right) \\

\end{align}\]

Using formula (1), the sum of (m + n) terms is equal to,

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

From equation (4), substituting \[\left( 2a+\left( m+n-1 \right)d \right)=0\] in the above equation, we get,

$\begin{align}

& {{S}_{m+n}}=\dfrac{\left( m+n \right)}{2}\left( 0 \right) \\

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

\end{align}$

Hence, we have proved that the sum of (m + n) terms of this A.P. is zero.

Note: In the equation (4), we can cancel the factor (m - n) only when $m-n\ne 0$. Since this question, m and n are written separately, we can assume that m and n are different and hence, we can say $m-n\ne 0$.

__Complete step-by-step answer:__Before proceeding with the question, we must know the formula that will be required to solve this question.

In arithmetic progression, if we have an arithmetic progression with its first term equal to a and its common difference equal to d, then, the sum of its first r terms is given by the formula,

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

In this question, it is given that the sum of m terms of an A.P. is the same as the sum of its n terms and we have to show that the sum of its (m + n) terms is zero.

Let us assume an A.P. with its first term as a and common difference as d. Using formula (1), the sum of m terms of this A.P. is equal to,

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

Using formula (1), the sum of n terms of this A.P. is equal to,

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ . . . . . . . . . . . . (3)

Since in the question, it is given ${{S}_{m}}={{S}_{n}}$, from equation (2) and (3), we get,

\[\begin{align}

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

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

& \Rightarrow 2am+{{m}^{2}}d-md=2an+{{n}^{2}}d-nd \\

& \Rightarrow 2am-2an+{{m}^{2}}d-{{n}^{2}}d-md+nd=0 \\

& \Rightarrow 2a\left( m-n \right)+\left( {{m}^{2}}-{{n}^{2}} \right)d-d\left( m-n \right)=0 \\

\end{align}\]

There is a formula ${{m}^{2}}-{{n}^{2}}=\left( m-n \right)\left( m+n \right)$. Substituting this formula in the above equation, we get,

\[\begin{align}

& 2a\left( m-n \right)+\left( m-n \right)\left( m+n \right)d-d\left( m-n \right)=0 \\

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

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

& \Rightarrow \left( 2a+\left( m+n-1 \right)d \right)=0..................\left( 4 \right) \\

\end{align}\]

Using formula (1), the sum of (m + n) terms is equal to,

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

From equation (4), substituting \[\left( 2a+\left( m+n-1 \right)d \right)=0\] in the above equation, we get,

$\begin{align}

& {{S}_{m+n}}=\dfrac{\left( m+n \right)}{2}\left( 0 \right) \\

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

\end{align}$

Hence, we have proved that the sum of (m + n) terms of this A.P. is zero.

Note: In the equation (4), we can cancel the factor (m - n) only when $m-n\ne 0$. Since this question, m and n are written separately, we can assume that m and n are different and hence, we can say $m-n\ne 0$.

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