Let the sum of n, 2n, and 3n terms of an A.P. be S₁, S₂, and S₃ respectively. Show that S₃ = 3(S₂ − S₁).

Let a and d be the first term and the common difference of the A.P. respectively.

We know that the sum of k terms of an A.P. is given by:

Sₖ = (k/2)[2a + (k − 1)d]

∴ S₁ = (n/2)[2a + (n − 1)d]     ...(1)

S₂ = (2n/2)[2a + (2n − 1)d] = n[2a + (2n − 1)d]     ...(2)

S₃ = (3n/2)[2a + (3n − 1)d]     ...(3)

From (1) and (2), we have:

S₂ − S₁ = n[2a + (2n − 1)d] − (n/2)[2a + (n − 1)d]

= n{(4a + 4nd − 2d − 2a − nd + d)/2}

= n[(2a + 3nd − d)/2]

= (n/2)[2a + (3n − 1)d]

Therefore,

3(S₂ − S₁) = (3n/2)[2a + (3n − 1)d] = S₃     [From (3)]

Hence, S₃ = 3(S₂ − S₁).