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

Answer

N/a

Explanation

Let the first term and common difference of A.P. be 'a' and 'd' respectively.
`"Now "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`
`rArr" "(m-n)[2a+(m+n-1)d]0`
`rArr " "2a+(m+n-1)d=0" ...(1) "(because mnen)`
`therefore" "S_(m+n)=(m+n)/(2)[2a+(m+n-1)d]`
`=(m+n)/(2)xx0" [From eq. (1)]"`
`=0.`