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If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)

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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 difference
Now,
`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)






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