if Sn denotes the sum of the first n terms of an AP ,Prove that S12=3(S8 -S4)
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s12=3 (s8-s4)
s12= 12/2 [2a+(12-1)d]
6 [2a+11d] 1
s8=8/2 [2a+(8-1)d]
4[2a+7d] 2
s4=4/2 [2a+(4-1)d]
2 [2a+3d] 3
keep eq.123 on its place
6 (2a+11d)=3 [4 (2a+7d)-2 (2a+3d)]
12a+66d=3 (8a+28d-4a-6d)
=3 (4a+22d)
=12a+66d
hence proved
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