If the sum of m terms of an A.P. is same as the sum of its n terms, then the sum of its (m+n) terms is

Options

A. mn
B. `-mn`
C. 1/mn
D. 0

Answer

D

Explanation

Let a be the first term and d be the common difference of the given A.P. Then
`S_(m)=S_(n)`
`rArrm/2[2a+(m-1)d]=n/2[2a+(n-1)d]`
or `2a(m-n)+{m(m-1)-n(n-1)}d=0`
or `2a(m-n)+{(m^(2)-n^(2))-(m-n)}d=0`
or `(m-n)[2a+(m+n-1)d]=0`
or `2a+(m+n-1)d=0` ["becausem-nne0"]
Now,
`S_(m+n)=(m+n)/2[2a+(m+n-1)d]=(m+n)/2xx0`=0
[Using 1]