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

Let sum of `n`, `2n`, `3n`, terms of an A.P are `S_(1), S_(2), S_(3)` respectively. Prove that `S_(3) = 3 (S_(2) - S_(1))`.

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Let a be the first term and d be the common difference of the given A.P. Then, sum of n terms is
`S_(1)=n/2[2a+(n-1)d]` (1)
Sum of 2n terms,
`S_(2)=`Sum of 2n terms=`(2n)/2[2a+(2n-1)d]` (2)
Sum of 3n terms
`S_(3)=(3n)/2[2a+(3n-1)d]` (3)
Now, `S_(2)-S_(1)=(2n)/2[2a+(2n-1)d]-n/2[2a+(n-1)d]`
`=n/2[2{2a+(2n-1)d}-{2a+(n-1)d}]`
`=n/2[2a+(3n-1)d]`
`therefore 3(S_(2)-S_(1))=(3n)/2[2a+(3n-1)d]=S_(3)` [Using (3)]
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