Question

If S_n denotes the sum of first n terms of an AP, then prove that S₁₂ = 3(S₈ – S₄).

Answer 1

Let a be first term and d be common difference of an AP

Then

= 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.