If S_n denotes the sum of first n terms of an A.P., prove that :

S₁₂ = 3(S₈ − S₄)

By formula,

$S_n = \dfrac{n}{2}[2a + (n - 1)d]$

where, a = first term, d = common difference

$\Rightarrow S${12} = \dfrac{12}{2}[2a + 11d] \\[1em] \Rightarrow S{12} = 6(2a + 11d) \\[1em] \Rightarrow S{12} = 12a + 66d \text{ ….(1)} \\[1em] \Rightarrow S{8} = \dfrac{8}{2}[2a + 7d] \\[1em] \Rightarrow S{8} = 4(2a + 7d) \\[1em] \Rightarrow S{8} = 8a + 28d \text{ ….(2)} \\[1em] \Rightarrow S{4} = \dfrac{4}{2}[2a + 3d] \\[1em] \Rightarrow S{4} = 2(2a + 3d) \\[1em] \Rightarrow S_{4} = 4a + 6d \text{ ….(3)}

Subtracting equation (3) from equation (2), we get :

⇒ S₈ - S₄ = (8a + 28d) - (4a + 6d)

⇒ S₈ - S₄ = 8a + 28d - 4a - 6d

⇒ S₈ - S₄ = 4a + 22d

Multiplying the above equation by 3, we get :

⇒ 3(S₈ - S₄) = 3(4a + 22d)

⇒ 3(S₈ - S₄) = 12a + 66d

⇒ 3(S₈ - S₄) = S₁₂

Hence, proved that S₁₂ = 3(S₈ − S₄).