Question

Sum of first n terms of an A.P. is m and sum of first m terms of same A.P. is n. Then prove that sum of first (m+n) terms of the A.P. is -(m+n).

Answer 1

the sum of first m terms of an AP
is n
m/2(2a+(m-1)d=n
2am +m(m-1)d=2n.....(1)

and the sum of first n terms of an AP is m
n/2(2a +(n-1)d=m
2an+n(n-1)d=2m..(2)

NOW SUBTRACT (1) AND (2)
we get
2am +m(m-1)d-(2an+n(n-1)d)=2n-2m
2a(m-n) +(m(square)-m-n(square)+n)d=-2(m-n)
2a(m-n)+((m-n) (m+n)-(m-n))d=-2(m-n)
TAKING (m-n) COMMON
we get
2a+(m+n-1)d=-2.....(3)
NOW
Sm+n=(m+n/2)(2a+(m+n-1)d
(m+n/2)(-2)
Sm+n=-(m+n).....(your answer)
HOPE IT WORKS