1.

The sum of n,2n,3n term of an AP are S1,S2,S3 respectively . Prove that S3=3(S2_S1)

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}


Discussion

No Comment Found