Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that
S3 = 3(S2 – S1)
Let a and b be the first term and the common difference of the A.P. respectively.
Therefore,
S1 = (n/2)[2a + (n - 1)d] …(1)
S2 = (2n/2)[2a + (2n - 1)d] …(2)
S3 = (3n/2)[2a + (3n - 1)d] …(3)
From (1) and (2), we obtain
S2 - S1 = (2n/2)[2a + (2n - 1)d] - (n/2)[2a + (n - 1)d]
= (n/2)[{4a + (4n - 2)d} - {2a + (n - 1)d}]
= (n/2)[4a + 4nd - 2d - 2a - nd + d]
= (n/2)[2a + 3nd - d]
= (1/3) × (3n/2)[2a + (3n - 1)d]
= (1/3)S3
Thus, S3 = 3(S2 - S1)
Hence, the given result is proved.