Question

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

Answer 1

Sum of n terms of an AP,

∵ S_n = `n/2[2a + (n - 1)d]`   ...(i)

∴ S₈ = `8/2[2a + (8 - 1)d]`

= 4(2a + 7d)

= 8a + 28d

And S₄ = `4/2[2a + (4 - 1)d]`

= 2(2a + 3d)

= 4a + 6d

Now, S₈ – S₄

= 8a + 28d – 4a – 6d

= 4a + 22d   ...(ii)

And S₁₂ = `12/2[2a + (12 - 1)d]`

= 6(2a + 11d)

= 3(4a + 22d)

= 3(S₈ – S₄)  ...[From equation (ii)]

∴ S₁₂ = 3(S₈ – S₄) 

Hence proved.