If S1,S2,S3 be the sum of n,2n,3n terms respectively of an A.P .Prove that S3 = 3(S2 - S1)
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use formula
S=n/2 {2a+(n-1) d}
S1=n/2 {2a+(n-1) d}
S2=2n/2 {2a+(n-1) d}
S3=3n/2 {2a+(n-1) d}
now,
RHS=3 (S2-S1)
=3 [2n/2{2a+(2n-1) d}-n/2 {2a+(n-1) d}]
=3 [n/2 {4a+4nd-2d-2a-nd+d}]
=3n/2 {2a +3nd-d}
=3n/2 {2a+(3n-1) d} =S3=LHS
S=n/2 {2a+(n-1) d}
S1=n/2 {2a+(n-1) d}
S2=2n/2 {2a+(n-1) d}
S3=3n/2 {2a+(n-1) d}
now,
RHS=3 (S2-S1)
=3 [2n/2{2a+(2n-1) d}-n/2 {2a+(n-1) d}]
=3 [n/2 {4a+4nd-2d-2a-nd+d}]
=3n/2 {2a +3nd-d}
=3n/2 {2a+(3n-1) d} =S3=LHS
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