Question

If S_n¹ denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ − S₄).

Answer 1

Let a and d be the first term and the common difference of AP, respectively.

We know

`S_n=n/2[2a+(n-1)d]`

`:.S_8=8/2[2a+(8-1)d] " and "S_4= 4/2[2a+(4-1)d]`

⇒S₈=4(2a+7d) and S₄=2(2a+3d)

⇒S₈=8a+28d and S₄=4a+6d

Now,

3(S₈−S₄)=3(8a+28d−4a−6d)

= 3(4a+22d)

= 6(2a+11d)

`=12/2[2a+(12-1)d]`

= S₁₂

∴ S₁₂=3(S₈−S₄)