If the sum of first `m` terms of an A.P. is the same as the sum of its first `n` terms, show that the sum of tis `(m+n)` terms is zero.

Explanation

Let the first term and common difference of A.P. be 'a' and 'd' respectively.
Given that, `S_(m)=S_(n)`
`rArr (m)/(2)[2a+(m-1)d]=(n)/(2)[2a+(n-1)d]`
`rArr 2am+m^(2)-m)d=2an+(n^(2)-n)d`
` rArr 2a(m-n)+[(m^(2)-m)-(n^(2)-n)]d=0`
`rArr 2a(m-n)+[(m^(2)-n^(2))-(m-n)]d=0`
`rArr 2a(m-n)+[(m-n)(m+n)-(m-n)]d=0`
`rArr 2a(m-n)+(m-n)(m+n-1)d=0`
`(m-n)[(2a+(m+n-1)d]=0`
`rArr 2a+(m+n-1)d=0 " " ...(1) " "(.:' m!=n)`
Now, `S_(m+n)=(m+n)/(2)[2a+(m+n-1)d]`
`=(m+n)/(2)xx0 " " ` [from eq. (1)]
`=0 " " ` Hence Proved.