The sum of n,2n,3n terms of an Ap.are S1,S2,S3 prove that S3 =3(S2-S1) give answer immeditely

The sum of n,2n,3n terms of an Ap.are S1,S2,S3 prove that S3 =3(S2-S1)

1.	The sum of n,2n,3n terms of an Ap.are S1,S2,S3 prove that S3 =3(S2-S1) give answer immeditely
Answer» {tex}{S_1} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex}{S_2} = \\frac{{2n}}{2}\\left[ {2a + (2n - 1)d} \\right]{/tex}{tex}{S_3} = \\frac{{3n}}{2}\\left[ {2a + (3n - 1)d} \\right]{/tex}R.H.S = 3(S2 - S1){tex} = 3\\left[ {\\frac{{2n}}{2}(2a + (2n - 1)d - \\frac{n}{2}(2a + (n - 1)d)} \\right]{/tex}{tex} = 3\\left[ {\\frac{n}{2}\\left[ {4a + 4nd - 2d - 2a - nd + d} \\right]} \\right]{/tex}{tex} = 3\\left[ {\\frac{n}{2}(2a + 3nd - d)} \\right]{/tex}{tex} = \\frac{{3n}}{2}\\left[ {2a + (3n - 1)d} \\right] = {S_3}{/tex}