If the sum of first m …

Let a be the first term and d be the common difference of the given A.P. Then,                     
S_m = S_n
{tex}\Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \}{/tex}
{tex} \Rightarrow{/tex} 2a (m - n) + {m (m - 1) - n (n - 1)} d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {m² - m - n² + n}d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {(m² - n²) - (m - n)} d = 0
{tex} \Rightarrow{/tex}2a (m - n) + {(m  - n) (m +n) - (m - n)} d = 0
{tex} \Rightarrow{/tex} (m - n) {2a + (m + n - 1)d} = 0
{tex} \Rightarrow{/tex} 2a + (m + n - 1)d = 0 {tex} [ \because m - n \neq 0 ]{/tex} ...(i)
Now, {tex} S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \} = \frac { m + n } { 2 } \times {/tex}0 = 0 [Using (i)]