If the sum of first n,2n,3n terms of the AP are S1,S2,S3 respectively,then prove that S3=3(S2-S1).

Answer 1

Sol:
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

Answer 2

Please help me how 4a +(4n-2)d come in the second step