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1 Pure Mathematics

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Question 256

An arithmetic series has first term a a\,a and common difference ddd. The sum of the first n n\,n terms of the series is SnS_nSn​.

By writing Sn S_n\,Sn​ out both forwards and in reverse, and adding the two expressions, prove that

Sn=12n[2a+(n−1)d]S_n = \tfrac{1}{2}n\left[2a + (n - 1)d\right]Sn​=21​n[2a+(n−1)d]

[5]
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