Question

The sum of n, 2n, and 3n terms of an AP are S1,S2,S3 . prove that 3(s2-

Answer 1

Let ‘a’ be the first term of the AP and ‘d’ be the common difference

S1 = (n/2)[2a + (n – 1)d] --- (1)

S2 = (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2)

S3 = (3n/2)[2a + (3n – 1)d] --- (3)

Consider the RHS: 3(S2 – S1)



 = S3

 = L.H.S

∴ S3 = 3(S2 - S1) .

Answer 2

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