If S_n denotes the sum of first n terms of an A.P., prove that S₃₀ = 3(S₂₀ - S₁₀).

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

We need to prove S₃₀ = 3(S₂₀ - S₁₀).

$\Rightarrow \text{L.H.S.} = S${30} = \dfrac{30}{2}[2a + (30 - 1)d] \\[1em] = 15[2a + 29d] \\[1em] = 30a + 335d \\[1em] \Rightarrow \text{R.H.S.} = 3(S{20} - S_{10}) \\[1em] = 3\Big(\dfrac{20}{2}[2a + (20 - 1)d] - \dfrac{10}{2}[2a + (10 - 1)d]\Big) \\[1em] = 3\Big(10[2a + 19d] - 5[2a + 9d]\Big) \\[1em] = 3(20a + 190d - 10a - 45d) \\[1em] = 3(10a + 145d) \\[1em] = 30a + 335d.

∴ L.H.S. = R.H.S. = 30a + 335d.

Hence, proved that S₃₀ = 3(S₂₀ - S₁₀).