# Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show thatS3 = 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.

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