If the sum of first m terms of an A.P. be n and sum of first n terms be m, then show that the sum of its first (m+n) terms is -(m+n).

Explanation

Let a be the first term and d be the common difference of the given AP. Then,
`S_(m) = n rArr (m)/(2) [2a + (m-1)d] =n`
`rArr 2am + m(m-1)d = 2n. " "…(i)`
`"And,"S_(n) = m rArr (n)/(2) [2a + (n-1)d] = m`
`rArr 2an + n(n-1)d = 2m. " "...(ii)`
On subtracting (ii) from (i), we get
`2a(m-n) + [(m^(2) -n^(2)) - (m-n)]d = 2(n-m)`
`rArr (m-n)[2a + (m + n-1)d] = 2(n-m)`
`rArr 2a + (m + n-1)d = -2 " "...(iii)`
Sum of the first (m+n) terms of the given AP
`= ((m+n))/(2) * {2a + (m + n-1)d}`
` = ((m+n))/(2) * (-2) = -(m+n) ["using (iii)"]`
Hence, the sum of first (m+n) terms of the given AP is -(m+n).