If S1,S2  and  S3  denote the sum of first n1,n2  and  n3  terms respectively of an AP, then S1n1(n2−n3)+S2n2(n3−n1)+S3n3(n1−n2)  is equal to

We have,S1=n12[2a+(n1−1)d]⇒2S1n1=2a+(n1−1)dS2=n22[2a+(n2−1)d]⇒2S2n2=2a+(n2−1)dS3=n32[2a+(n3−1)d]⇒2S3n3=2a+(n3−1)d now 2S1n1(n2−n3)+2S2n2(n3−n1)+2S3n3(n1−n2)=[2a+(n1−1)d](n2−n3)+[2a+(n2−1)d](n3−n1)  +[2a+(n3−1)d](n1−n2)=2a-dn2-n3+n3-n1+n1-n2+dn1n2-n3+n2n3-n1+n3n1-n2=0