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Let the sum of n, 2n, 3n terms of an A.P...

Let the sum of n, 2n, 3n terms of an A.P. be `S_1,S_2`and `S_3`, respectively, show that `S_3=3(S_2-S_1)`.

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Here, `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, `S_2 - S_1 = n/2(4a+(4n-2)d - 2a-(n-1)d)`
`= n/2(2a+(3n-1)d)`
So, `3(S_2-S_1) = 3**n/2(2a+(3n-1)d) =(3n)/2(2a+(3n-1)d) = S_3`
So, `3(S_2-S_1) = S_3`
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