Proof for Arithmetic Series

It is a simple proof for Arithmetic Progression, and one should memorise it.

Terms: \(a, a+d, a+d,..., a+(n-2)d, a+(n-1)d\)

Now set up two opposite direction sums

Sum1: \(S_n=(a)+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d)\)

Sum2: \(S_n=(a+(n-1)d)+(a+(n-2)d)+...+(a+2d)+(a+d)+(a)\)

Sum1 + Sum2 results in

\(2S_n=(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+\newline\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...+(2a+(n-1)d)+(2a+(n-1)d)\)

\(\therefore\)

\(2S_n=n[2a+(n-1)d]\)

\(S_n=\frac{n}{2}\left( 2a+(n-1)d\right) =\frac{n}{2}\left( a+\color{red} a+(n-1)d\right)\)

Where the red term is the last term, therefore we have

\(S_n=\frac{n}{2}\left( a+\color{red}last\,term\right)\,\,\,\,\,\,\,\square\)