If Sn denotes the sum of first n terms of an AP, then prove that S12 = 3(S8 – S4).
Let a be first term and d be common difference of an AP
Then
Taking LHS
= 6(2a + (n - 1)d
= (12 - 6)(2a + (n - 1)d)
= 3(4 - 2)(2a + (n - 1)d)
= 3[ (4 - 2)(2a + (n - 1)d)]
= 3[ 4(2a + (n - 1)d) - 2(2a + (n - 1)d)]
= 3(S8 - S4) [ By eqn 1 & eqn 2]
= RHS
Hence proved.