if Sn denotes the sum of first n terms of an A.P., prove that S12=3(S8-S4).

On Equating the RHS we get ,

3(S8 - S4)

=> 3 { n/2 * [ 2a + (n-1)d ] - n/2 * [2a + (n-1)d ] }

=> Substituting 8 and 4 in n we get,

=> 3 { 8/2 * [ 2a + (8-1)d ] - 4/2 * [2a + (4-1)d ] }

=> 3 { 4 [2a + 7d] - 2[2a + 3d] }

=> 3 { 8a +28d - 4a - 6d }

=> 3 {4a +22d }

=> 6 { 2a  + 11d }                                                                  ( Taking 2 common}

=> 12/2 ( 2a + (12-1)d }                                                       (Rearranging the terms)

=> S12