Question

SN denotes the sum of the first n terms of AP prove that S12 is equals to 3 (s 8 - S4)

Answer 1

Sn=n/2[2a+(n-1)d]
S12=6(2a+11d)
S8=4(2a+7d)
S4=2(2a+3d)
Now
3(S8+S4)
=3(8a+28d-4a-6d)
=3(4a+22d)
=6(2a+11d)
=S12
Hence proved
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Answer 2

S12=n/2[2a+(n-1)d]

S₁₂=3[S₈-S₄]

n/2[2a+(n-1)d]=3[n/2{2a+(n-1)d}]-[n/2{2a+(n-1)d}]

6[2a+(n-1)d]=3[8a+4nd-4d-4a-2nd+2d]

12a+6nd-6d=3[4a+2nd-2d]

12a+6nd-6d=12a+6nd-6d

LHS=RHS

                           Hence Proved