if sn denotes the sum of first n terms of an ap prove that s12= 3(s8-s4)
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Step-by-step explanation:
Sn= (n/2)×{2a+(n-1)d}
S8=4×{2a+7d}=8a+28d
S4=2×{2a+3d}=4a+6d
L.H.S.:
S12=6×{2a+11d}= 3×{4a+22d}=3×{(8a+28d)-(4a+6d)}=3×(S8-S4).....(R.H.S) (proved)
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