Question

Suppose the sum of the dirst m teams of a arithmetic progression is n and the sum of its first n terms is m, where `mnen.` Then the dum of the first (m + n) terms of the arithmetic progression is
A. 1- mn
B. mn- 5
C. `-(m = n)`
D. m + n

Answer 1

Correct Answer - C
`S_(m)=m/2[2a+(m-1)d]=n" …(1)"`
`S_(n)=n/2[2a+(n-1)d]=m" …(2)"`
by (1) and (2)
`(m-n)a+(m-n){m+n-1}d/2= -(m-n)`
`rArr2a+(m+n-1)d= -2(mnen)`
`rArrS_(m+n)=(m+n)/2[2a+(m+n-1)d]= -(m+n)`