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Chandrashekhar Wankhede 7 years, 9 months ago
Sn = (n/2)(2a+(n-1)d)
S12= (12/2)(2a+(12-1)d)=12a+66d
Similarly
S8= (8/2)(2a+(8-1)d)= 8a+28d
S4= (4/2)(2a+(4-1)d) = 4a+6d
RHS = 3(S8-S4)= 3(8a+28d-4a-6d)
RHS =3(4a+22d) = 12a + 66d = LHS
Hence Proved
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