Question

if sn denotes the sum of 1st n terms of ap.prove that s12=3(s8-s4)

Answer 1

let a is the first term of Ap and d is the common difference
Sn=n/2 {2a+(n-1) d

now S12=12/2 {2a+(12-1) d}=12a+66d
S8=8/2 {2a+7d}=8a+28d
S4=4/2 {2a+3d}=4a+6d
LHS=S12=12a+66d
RHS=3 (S8-S4)=3 (8a+28d-4a-6d)=12a+66d
LHS =RHS

Answer 2

LHS= S12
=n/2 [2a+(n-1)d]
= 12/2 [ 2a+ (12-1)d]
= 6(2a+11d)
LHS = 12a+66d
RHS= 3(S8-S4)
= 3 [4(2a+7d) - 2(2a+3d)
=3(8a+28d-4a-6d)
= 3(4a+22d)
RHS = 12a+66d
therefore, LHS=RHS
Hence proved !