If the sum of first \'m\' terms of an AP is n,then show that the sum of its first (m+n)=-(m+n)

If the sum of first \'m\' terms of an AP is n,then show that the sum o

1.	If the sum of first \'m\' terms of an AP is n,then show that the sum of its first (m+n)=-(m+n)
Answer» Let a be the first term and d be the common difference of the given AP. Then,Sm = n\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{m}{2}{/tex}[2a + (m-1)d] = n{tex}\\Rightarrow{/tex}\xa02am + m(m- 1)d - 2n\xa0...... (i)And, Sn\xa0= m\xa0{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a + (n - 1)d] = m{tex}\\Rightarrow{/tex}\xa02an + n(n - 1)d = 2m ...... (ii)On subtracting (ii) from (i), we get2a(m-n) + [(m2 - n2) - (m - n)]d = 2(n - m){tex}\\Rightarrow{/tex}\xa0(m - n)[2a + (m + n - 1)d] = 2(n - m){tex}\\Rightarrow{/tex}\xa02a + (m + n- 1)d = -2 ..... (iii)Sum of the first (m + n) terms of the given AP=\xa0{tex}\\frac{{(m + n)}}{2}{/tex}{tex}\\cdot{/tex}{2a + (m + n - 1)d}{tex}= \\frac { ( m + n ) } { 2 } \\cdot ( - 2 ) = - ( m + n ){/tex}\xa0[using (iii)].Hence, the sum of first (m + n) terms of the given AP is -(m + n).