First term (a), second (b), last (c).

(d=b-a), number of terms (n): [ c=a+(n-1)d=a+(n-1)(b-a)\implies n-1=\dfrac{c-a}{b-a}\implies n=\dfrac{c-a}{b-a}+1=\dfrac{b+c-2a}{b-a} ]

[ S=\dfrac{n}{2}(a+c)=\dfrac{b+c-2a}{2(b-a)}(a+c)=\dfrac{(b+c-2a)(a+c)}{2(b-a)} ]

Proved.