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 the first term and common difference of the A.P. be 'a' and 'd' respectively.
Given that, `S_(m)=n`
`rArr (m)/(2)[2a+(m-1)d]=n`
`rArr 2am +(m^(2)-m)d=2n " " ` ...(1)
and `S_(n)=m`
`rArr (n)/(2)[2a+(n-1)d]=m`
`rArr 2an +(n^(2)-n)d=2m " " ...(2)`
Subtract eq. (2) from eq. (1) , we get
`2a(m-n)+{(m^(2)-n^(2))-(m-n)}d=2(n-m)`
`rArr 2a(m-n)+{(m-n)(m+n)-(m-n)}d=-2(m-n)`
`rArr 2a(m-n)+(m-n)(m+n-1)d=-2(m-n)`
`rArr 2a+(m+n-1)d=-2 " " ...(3)`
Now, `S_(m+n)=(m+n)/(2){2a+(m+n-1)d}`
`=(m+n)/(2)(-2)=-(m+n) " " `Hence Proved.