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 Solution If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)Solution:Your problem -> If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)Let a be the first term and d be the common differenceNow,S_1= n/2 [ 2a + (n - 1) d ]S_2 = (2n)/2 [ 2a + (2n - 1) d ]S_3 = (3n)/2 [ 2a + (3n - 1) d ]Now, 3(S_2 - S_1)= 3 [ (2n)/2 ( 2a + (2n - 1) d ) - n/2 ( 2a + (n - 1) d ) ]= 3 [ n/2 ( 2(2a) - 2a ) + n/2 ( 2(2n - 1) d - (n - 1) d ) ]= 3 [ n/2 ( 2a ) + n/2 ( 4n - 2 - n + 1) d ) ]= 3 [ n/2 ( 2a ) + n/2 ( 3n - 1) d ) ]= (3n)/2 [ 2a + ( 3n - 1) d ]= S_3 (Proved) Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me