if Sn denotes sum of n terms then prove that S12=3(S8 - S4)

1.	if Sn denotes sum of n terms then prove that S12=3(S8 - S4)
Answer» Let a be the first term and the common difference be d.Sn\xa0=\xa0{tex}\\frac n2{/tex}[2a + (n - 1)d]S12 = {tex}\\frac {12}2{/tex}[2a + (12 - 1)d]= 6[2a + 11d]= 12a + 66dS8 =\xa0{tex}\\frac 82{/tex}[2a + (8 - 1)d]= 4[2a + 7d]= 8a + 28dS4 =\xa0{tex}\\frac 42{/tex}[2a + (4- 1)d]= 2[2a + 3d]= 4a + 6d3(S8 - S4) = 3[(8a + 28d) - (4a + 6d)]= 3[8a + 28d - 4a - 6d]= 3[4a + 22d]= 12a + 66d= S12

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