Here i brought u challenging question …

Let a be the first term and d be the common difference of the given AP. Then,
S_m = n {tex}\Rightarrow{/tex} {tex}\frac{m}{2}{/tex}[2a + (m-1)d] = n
{tex}\Rightarrow{/tex} 2am + m(m- 1)d - 2n ...... (i)
And, S_n = m {tex}\Rightarrow{/tex}{tex}\frac{n}{2}{/tex}[2a + (n - 1)d] = m
{tex}\Rightarrow{/tex} 2an + n(n - 1)d = 2m ...... (ii)
On subtracting (ii) from (i), we get
2a(m-n) + [(m² - n²) - (m - n)]d = 2(n - m)
{tex}\Rightarrow{/tex} (m - n)[2a + (m + n - 1)d] = 2(n - m)
{tex}\Rightarrow{/tex} 2a + (m + n- 1)d = -2 ..... (iii)
Sum of the first (m + n) terms of the given AP
= {tex}\frac{{(m + n)}}{2}{/tex}{tex}\cdot{/tex}{2a + (m + n - 1)d}
{tex}= \frac { ( m + n ) } { 2 } \cdot ( - 2 ) = - ( m + n ){/tex} [using (iii)].
Hence, the sum of first (m + n) terms of the given AP is -(m + n).