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Question
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Solution
Sum of n terms of an AP,
∵ Sn = `n/2[2a + (n - 1)d]` ...(i)
∴ S8 = `8/2[2a + (8 - 1)d]`
= 4(2a + 7d)
= 8a + 28d
And S4 = `4/2[2a + (4 - 1)d]`
= 2(2a + 3d)
= 4a + 6d
Now, S8 – S4
= 8a + 28d – 4a – 6d
= 4a + 22d ...(ii)
And S12 = `12/2[2a + (12 - 1)d]`
= 6(2a + 11d)
= 3(4a + 22d)
= 3(S8 – S4) ...[From equation (ii)]
∴ S12 = 3(S8 – S4)
Hence proved.
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