Question

If the sum of `n , 2n , 3n` terms of an AP are `S_1,S_2,S_3` respectively . Prove that `S_3=3(S_2-S_1)`

Answer 1

Let a be the first term and d be the common difference of the given AP. Then,
`S_(1) `=sum of n terms of the given AP,
`S_(2)` = sum of first 2n terms of the given AP,
`S_(3)` = sum of first 3n terms of the given AP.
`therefore S_(1) = (n)/(2) * {2a + (n-1)d}, S_(2) = (2n)/(2) * {2a + (2n-1)d}`, and
`S_(3) = (3n)/(2) * {2a + (3n-1)d}`
`rArr 3(S_(2) -S_(1)) = 3 * [{2na + n (2n-1)d} -{na + (1)/(2)n(n-1)d}]`
`=3* [na + (3)/(2)n^(2)d -(1)/(2)nd] = (3n)/(2) * [2a +3nd-d]`
`= (3n)/(2) * [2a + (3n-1)d} =S_(3)`
Hence, `S_(3) = 3(S_(2)-S_(1))`