If in an A.P. the sum …

Let a be the first term and d be the common difference of the given A.P. Then,
S_m = n
{tex} \Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = n{/tex}
{tex}\Rightarrow{/tex} 2am + m (m - 1) d = 2n ...(i)
and, S_n = m
{tex}\Rightarrow \quad \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \} = m{/tex}
{tex}\Rightarrow{/tex} 2an + n (n -1)d = 2m ...(ii)
Subtracting equation (ii) from equation (i), we get
2a (m - n) + {m (m -1) - n (n - 1)} d = 2n - 2m
{tex}\Rightarrow{/tex} 2a (m - n) + {(m² - n²) - (m - n)} d = -2 (m - n)
{tex}\Rightarrow{/tex} 2a + (m + n -1) d = - 2 [On dividing both sides by (m - n)] ...(iii)
Now, {tex}S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \}{/tex}
{tex}\Rightarrow \quad S _ { m + n } = \frac { ( m + n ) } { 2 } ( - 2 ){/tex} [Using (iii)]
{tex}\Rightarrow{/tex} S_m+n = - (m + n)