If sum of m terms of an AP is equal to sum of n terms of an AP. Then prove that Sum of n+m terms = 0

If sum of m terms of an AP is equal to sum of n terms of an AP. Then p

1.	If sum of m terms of an AP is equal to sum of n terms of an AP. Then prove that Sum of n+m terms = 0
Answer» Let a be the first term and d be the common difference of the given A.P. Then, Sm = Sn{tex}\\Rightarrow \\quad \\frac { m } { 2 } \\{ 2 a + ( m - 1 ) d \\} = \\frac { n } { 2 } \\{ 2 a + ( n - 1 ) d \\}{/tex}{tex} \\Rightarrow{/tex}\xa02a (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}\xa02a (m - n) + {(m2 - n2) - (m - n)} d = 0{tex} \\Rightarrow{/tex}2a (m - n) + {(m\xa0 - n) (m +n) - (m - n)} d = 0{tex} \\Rightarrow{/tex}\xa0(m - n) {2a + (m + n - 1)d} = 0{tex} \\Rightarrow{/tex}\xa02a + (m + n - 1)d = 0\xa0{tex} [ \\because m - n \\neq 0 ]{/tex}\xa0...(i)Now,\xa0{tex} S _ { m + n } = \\frac { m + n } { 2 } \\{ 2 a + ( m + n - 1 ) d \\} = \\frac { m + n } { 2 } \\times {/tex}0 = 0 [Using\xa0(i)]