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Question
If Sn1 denotes the sum of first n terms of an A.P., prove that S12 = 3(S8 − S4).
Solution
Let a and d be the first term and the common difference of AP, respectively.
We know
`S_n=n/2[2a+(n-1)d]`
`:.S_8=8/2[2a+(8-1)d] " and "S_4= 4/2[2a+(4-1)d]`
⇒S8=4(2a+7d) and S4=2(2a+3d)
⇒S8=8a+28d and S4=4a+6d
Now,
3(S8−S4)=3(8a+28d−4a−6d)
= 3(4a+22d)
= 6(2a+11d)
`=12/2[2a+(12-1)d]`
= S12
∴ S12=3(S8−S4)
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