if the sum m term ap is same as the sum of its n term show that the sum of m+m term is zero
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m/2(2a+(m-1)d)=n/2(2a+(n-1)d)
m(2a+(m-1)d)=n(2a+(n-1)d)
2am+dm^2-md=2an+dn^2-nd
2a(m-n)+d(m^2-n^2)-d(m-n)=0
(m-n)(2a+dm+dn-d)=0
m=n
or
2a+d(m+n-1)=0
As m not equal to n,
2a+d(m+n-1)=0
so
(m+n)/2(2a+(m+n-1)d)=0