S3=3(S2-S1)

1.	S3=3(S2-S1)
Answer» Q :\xa0The sum of n, 2n ,3n terms of an AP are S1,S2,S3,respectively. Prove that S3 =3(S2- S1)Answer:Let ‘a’ be the first term of the AP and ‘d’ be the common differenceS1\xa0= (n/2)[2a + (n – 1)d] --- (1)S2\xa0= (2n/2)[2a + (2n – 1)d] = n[2a + (n – 1)d] --- (2)S3\xa0= (3n/2)[2a + (3n – 1)d] --- (3)Consider the RHS: 3(S2\xa0– S1)\xa0= S3\xa0= L.H.S∴ S3\xa0= 3(S2\xa0- S1) .

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